Mathematical Reflections 2017, Issue 4 - Problem S416

Problem:
Let $f:\mathbb{N} \to \{\pm 1\}$ be a function such that $f(mn)=f(m)f(n)$ for all $m,n \in \mathbb{N}$. Prove that
there are infinitely many $n$ such that $f(n)=f(n+1)$.

Proposed by Oleksi Klurman, University College London, UK

Solution:
Let $n \in \mathbb{N}$. As $f$ is completely multiplicative, then $f(n^2)=(f(n))^2=1$. If there are infinitely many $n$ such that $f(n^2-1)=1$, we are done. So, assume that there are only finitely many $n$ such that $f(n^2-1)=1$. Then, there are infinitely many $n$ such that $f(n^2-1)=-1$. Since $f(n^2-1)=f(n-1)f(n+1)$, then there are infinitely many pairs $\{f(n-1),f(n+1)\}=\{-1,1\}$. Since $f(n)=\pm 1$, it follows that there are infinitely many $n$ such that $f(n)=f(n+1)$.

Mathematical Reflections 2017, Issue 4 - Problem S415

Problem:
Let $$f(x)=\dfrac{(2x-1)6^x}{2^{2x-1}+3^{2x-1}}.$$
Evaluate $$f\left(\dfrac{1}{2018}\right)+f\left(\dfrac{3}{2018}\right)+\ldots+f\left(\dfrac{2017}{2018}\right).$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Observe that $$f(1-x)=\dfrac{(1-2x)6^{1-x}}{2^{1-2x}+3^{1-2x}}=\dfrac{(1-2x)6^x}{2^{2x-1}+3^{2x-1}}=-f(x),$$ i.e. $f(x)+f(1-x)=0$.
So,
$$\begin{eqnarray*} \sum_{k=1}^{1009} f\left(\dfrac{2k-1}{2018}\right)&=&\sum_{k=1}^{504}\left(f\left(\dfrac{2k-1}{2018}\right)+f\left(\dfrac{2018-(2k-1)}{2018}\right)\right)+f\left(\dfrac{1009}{2018}\right)\\&=&f\left(\dfrac{1009}{2018}\right)\\&=&f\left(\dfrac{1}{2}\right)\\&=&0. \end{eqnarray*}$$

Mathematical Reflections 2017, Issue 4 - Problem J417

Problem:
Solve in positive real numbers the equation $$\dfrac{x^2+y^2}{1+xy}=\sqrt{2-\dfrac{1}{xy}}.$$

Proposed by Adrian Andreescu, Dallas, Texas, USA

Solution:
Let $s=x+y$ and $p=xy$. Then, the given equation can be written as $$\dfrac{s^2-2p}{1+p}=\sqrt{2-\dfrac{1}{p}}.$$
By the AM-GM Inequality, we have $s^2-2p \geq 2p$, so $$\sqrt{2-\dfrac{1}{p}} \geq \dfrac{2p}{1+p} \iff \dfrac{(p-1)^2(2p+1)}{p(p+1)^2} \leq 0.$$ Since $p \geq 0$, it follows that $p=1$, which gives $s=2$. So, $x+y=2$ and $xy=1$, which yields $(x,y)=(1,1)$.

Mathematical Reflections 2017, Issue 4 - Problem J415

Problem:
Prove that for all real numbers $x,y,z$ at least one of the numbers
$$2^{3x-y}+2^{3x-z}-2^{y+z+1}$$ $$2^{3y-z}+2^{3y-x}-2^{z+x+1}$$ $$2^{3z-x}+2^{3z-y}-2^{x+y+1}$$
is nonnegative.

Proposed by Adrian Andreescu, Dallas, Texas, USA

Solution:
Let $a=2^x$, $b=2^y$, $c=2^z$. Adding the three given numbers, we get
$$\begin{array}{lll} S&=&\left(\dfrac{a^3}{b}+\dfrac{a^3}{c}-2bc\right)+\left(\dfrac{b^3}{c}+\dfrac{b^3}{a}-2ca\right)+\left(\dfrac{c^3}{a}+\dfrac{c^3}{b}-2ab\right)\\&=&\left(\dfrac{a^3}{b}+\dfrac{b^3}{a}-2ab\right)+\left(\dfrac{b^3}{c}+\dfrac{c^3}{b}-2bc\right)+\left(\dfrac{c^3}{a}+\dfrac{a^3}{c}-2ca\right). \end{array}$$
By the AM-GM Inequality, we get
$$\dfrac{a^3}{b}+\dfrac{b^3}{a} \geq 2ab, \qquad \dfrac{b^3}{c}+\dfrac{c^3}{b} \geq 2bc, \qquad \dfrac{c^3}{a}+\dfrac{a^3}{c} \geq 2ca.$$
So, $S \geq 0$ and we conclude that at least one of the three given numbers is nonnegative.

Mathematical Reflections 2017, Issue 3 - Problem O414

Problem:
Characterize all positive integers $n$ with the following property: for any two coprime divisors $a<b$ of $n$, $b-a+1$ is also a divisor of $n$.

Proposed by Vlad Matei, University of Wisconsin, Madison, USA

Solution:
It's easy to see that if $n=p^k$, where $p$ is a prime and $k \in \mathbb{Z}^+$, then $n$ satisfies the condition. Assume that $n$ has at least two prime divisors. Let $n=mp^k$, where $p$ is the smallest prime dividing $n$ and $(m,p)=1$. Clearly $p<m$ and both $p$ and $m$ are divisors of $n$. If $n$ satisfies the condition, then $m-p+1$ is also a divisor of $n$. Let $q$ be a prime such that $q \mid m$. Since $m-q<m-p+1<m$, then $q$ doesn't divide $m-p+1$. It follows that the only prime factor of $m-p+1$ is $p$. Hence, $m-p+1=p^a$ for some positive integer $a \leq k$, i.e. $$m=p^a+p-1.$$ Assume that $a \geq 2$. Since $p^{a-1} \mid n$ and $p^{a-1}<m$, then $m-p^{a-1}+1=p(p^{a-1}-p^{a-2}+1)$ is also a divisor of $n$. As $p^{a-1}-p^{a-2}+1$ is not divisible by $p$, then it must divide $m$. As
$$m=p^a+p-1=(p+1)(p^{a-1}-p^{a-2}+1)+p^{a-2}-2,$$ then $(p^{a-1}-p^{a-2}+1) \mid m$ if and only if $(p^{a-1}-p^{a-2}+1) \mid (p^{a-2}-2)$. We have $$p^{a-1}-p^{a-2}+1>p^{a-2}-2 \iff p^{a-2}(p-2)+3>0,$$ so there are no solutions if $a \geq 2$. If $a=1$, we get $m=2p-1$.

(i) If $k \geq 2$, then $p^2 \mid n$ and $p^2>2p-1$, so $p^2-(2p-1)+1=p^2-2p+2$ is also a divisor of $n$. If $p>2$, then $p^2-2p+2$ is not divisible by $p$, which gives $(p^2-2p+2) \mid m$, i.e. $(p^2-2p+2) \mid (2p-1)$ and this is true only if $p=3$. So, $m=5$ and $n=5\cdot 3^k$. Since $3^k>5$, then also $3^k-5+1=3^k-4$ is a divisor of $n$, which forces $3^k-4=5$, i.e. $k=2$ and $n=45$. If $p=2$, then $m=3$ and $n=3\cdot 2^k$. Since $2^k>3$, then also $2^k-3+1=2(2^{k-1}-1)$ is a divisor of $n$ and this implies $k=2$ or $k=3$. An easy check shows that indeed $n \in \{12,24\}$.

(ii) If $k=1$, then $n=(2p-1)p$. Let $q$ be the smallest prime divisor of $2p-1$. Then $q \mid n$, $p<q$ and so $q-p+1$ is a divisor of $n$. So, $(q-p+1) \mid (2p-1)$ or $(q-p+1) \mid p$. In the first case, by the minimality of $q$, it must be $q<q-p+1$, contradiction. In the second case, $q-p+1=1$ or $q-p+1=p$, which gives $q=2p-1$.

In conclusion, we get $n=p^k$, where $k \in \mathbb{Z}^+$, or $n=(2p-1)p$, where $p$ is prime and $2p-1$ is prime, or $n \in \{12,24,45\}$.

Mathematical Reflections 2017, Issue 3 - Problem O410

Problem:
On each cell of a chess board it is written a number equal to the amount of the rectangles that contain this cell. Find the sum of all the numbers.

Proposed by Robert Bosch, USA 

Solution:
Consider an $n \times n$ chessboard and let $S$ be the required sum. Observe that writing on each cell the number equal to the amount of the rectangles containing the cell is equivalent to perform the following operation on the chessboard: at the beginning write zero on each cell of the chessboard, then for each $i \times j$ rectangle ($1 \leq i,j \leq n$) add $1$ to all its cells. So, we only have to find the total number of rectangles, each counted with the number its of cells. The number of $i \times j$ rectangles is $(n+1-i)(n+1-j)$, where $1 \leq i,j \leq n$. Since each $i \times j$ rectangle contains $ij$ cells, then we have
$$\begin{eqnarray*} S&=& \sum_{i=1}^n \sum_{j=1}^n ij(n+1-i)(n+1-j) \\ &=& \sum_{i=1}^n i(n+1-i) \sum_{j=1}^n j(n+1-j) \\ &=& \left(\sum_{k=1}^n k(n+1-k)\right)^2 \\ &=& {n+2 \choose 3}^2. \end{eqnarray*}$$

Mathematical Reflections 2017, Issue 3 - Problem O409

Problem:
Find all positive integers $n$ for which there are $n+1$ digits in base $10$, not necessarily distinct, such that at least $2n$ permutations of those digits produce $(n+1)$-digit perfect squares, with leading zeros not allowed. Note that two different permutations are considered distinct even if they lead to the same digit string due to repetition among the digits.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
We prove that all $n \geq 2$ satisfy the given property. Clearly, $n=1$ doesn't satisfy the given property. If $n=2$, take $144$ and the permutations $\{\textrm{id},(23),(13),(132)\}$ acting on the digits of $144$. If $n=3$, take $1444$ and the permutations $\{\textrm{id},(23),(24),(34),(234),(243)\}$ acting on the digits of $1444$. If $n=5$, take $160000$ and the $4!=24$ permutations acting on the digits of $160000$ moving only the zeros. Now, let $n \geq 4$ be even. Then, $n=2k$ for some integer $k \geq 2$. Observe that there are at least $(2k)!$ permutations acting on the digits of $10^{2k}$ that produces a perfect square (namely, the ones fixing $1$ in the first position and moving the other zeros) and $(2k)! \geq 4k$ for any integer $k \geq 2$. Let $n \geq 7$ be odd. Then, $n=2k-1$ for some integer $k \geq 3$. Observe that there are at least $k!\cdot (k-1)!$ permutations acting on the digits of $\underbrace{11\ldots 11}_{k \textrm{ ones}} \underbrace{55\ldots 55}_{k-1 \textrm{ fives}} 6$ that produces a perfect square and $k! \cdot (k-1)! \geq 2(2k-1)$ for any integer $k \geq 4$. The conclusion follows.

Mathematical Reflections 2017, Issue 3 - Problem U414

Problem:
Let $p < q < 1$ be positive real numbers. Find all functions $f: \mathbb{R} \to \mathbb{R}$ which satisfy the
conditions:

(i) $f(px+f(x))=qf(x)$ for all real numbers $x$,
(ii) $\lim_{x \to 0} \dfrac{f(x)}{x}$ exists and its finite.

Proposed by Florin Stanescu, Gaesti, Romania

Solution:
Clearly, $f(x)=0$ for all $x \in \mathbb{R}$ is a solution to the problem. Let $f \neq 0$. Observe that by condition (ii) it must be $f(0)=0$. We have two cases.

(i) $\displaystyle\lim_{x \to 0} \dfrac{f(x)}{x}=\ell \in \mathbb{R}\setminus\{0\}$. Then, from condition (i), we get
$$\dfrac{f(px+f(x))}{px+f(x)}=\dfrac{qf(x)}{x(p+\frac{f(x)}{x})}$$ and if $x \to 0$ we get $$\ell=\dfrac{q \ell}{p+\ell} \iff \ell=q-p.$$ So, $f(x)=(q-p)x$ is a solution to the problem.

(ii) $\displaystyle\lim_{x \to 0} \dfrac{f(x)}{x}=0$. Then, $f(x) \sim Cx^{1+\delta}$ if $x \to 0$, where $C \neq 0$ and $\delta>0$.
Then, from condition (i), we get
$$\dfrac{f(px+f(x))}{(px+f(x))^{1+\delta}}=\dfrac{qf(x)}{x^{1+\delta}(p+\frac{f(x)}{x})^{1+\delta}}$$ and if $x \to 0$ we get
$$C=\dfrac{qC}{p^{1+\delta}} \iff p^{1+\delta}=q,$$ contradiction.

We conclude that the only solutions are $f(x)=0$ and $f(x)=(q-p)x$.

Mathematical Reflections 2017, Issue 3 - Problem U412

Problem:
Let $P(x)$ be a monic polynomial with real coefficients, of degree $n$, which has $n$ real roots. Prove that if
$$P(c) \leq \left(\dfrac{b^2}{a}\right)^n,$$
then $P(ax^2+2bx+c)$ has at least one real root.

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
If $P(x)$ is an $n$-th degree monic polynomial with real coefficients and with $n$ real roots $\alpha_1,\ldots,\alpha_n$, then
$$P(x)=(x-\alpha_1)(x-\alpha_2)\cdot \ldots \cdot (x-\alpha_n).$$
Hence, $$P(c)=(c-\alpha_1)(c-\alpha_2)\cdot \ldots \cdot (c-\alpha_n) \leq \left(\dfrac{b^2}{a}\right)^n$$
and $$P(ax^2+2bx+c)=(ax^2+2bx+c-\alpha_1)(ax^2+2bx+c-\alpha_2)\cdot \ldots \cdot (ax^2+2bx+c-\alpha_n).$$
Assume by contradiction that $P(ax^2+2bx+c)$ has no real roots. Then, each factor has negative discriminant, i.e.
$$\begin{array}{lll} b^2-a(c-\alpha_1) & < & 0 \\ b^2-a(c-\alpha_2) & < & 0 \\ \vdots & \vdots & \vdots \\ b^2-a(c-\alpha_n) & < & 0. \end{array} \iff \begin{array}{lll} b^2 & < & a(c-\alpha_1) \\ b^2 & < & a(c-\alpha_2) \\ \vdots & \vdots & \vdots \\ b^2 & < & a(c-\alpha_n). \end{array}$$
Multiplying side by side all these inequalities, we get $(b^2)^n < a^n(c-\alpha_1)(c-\alpha_2)\cdot \ldots \cdot (c-\alpha_n)$, i.e.
$$\left(\dfrac{b^2}{a}\right)^n < (c-\alpha_1)(c-\alpha_2)\cdot \ldots \cdot (c-\alpha_n),$$ contradiction.

Mathematical Reflections 2017, Issue 3 - Problem U411

Problem:
Let $e$ be a positive integer. We say that a positive integer $m$ is awesome if $m$ has $\omega^e(m)$ digits in base ten, where $\omega(m)$ is the number of distinct primes that divide $m$. Prove that there are finitely many awesome numbers.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Let $\mathbb{P}$ be the set of primes and let
$$\mathcal{E}_n=\left\{\prod_{j=1}^{n^e} p_j^{k_j} : p_j \in \mathbb{P}, k_j \in \mathbb{N}^{\ast} \right\}$$ where the $p_j$ are distinct primes. A positive integer $m$ is awesome if and only if the two following conditions are simultaneously satisfied:

    (a) $m \in \mathcal{E}_n$
    (b) $10^{n^e-1} \leq m < 10^{n^e}$

for some $n \in \mathbb{N}^{\ast}$.
Let $$\mathcal{A}_n=\{m \in \mathcal{E}_n : 10^{n^e-1} \leq m < 10^{n^e}\}$$ be the set of awesome numbers with $n^e$ digits and let $$\mathcal{A}=\bigcup_{n \in \mathbb{N}^*} \mathcal{A}_n.$$ It's easy to see that $\mathcal{A}$ contains all the awesome numbers. \\
In order to prove that $\mathcal{A}$ is finite, we need to prove that there are finitely many nonempty $\mathcal{A}_n$.
We prove that for any fixed exponent $e$, there exists $k \in \mathbb{N}$ such that $\forall n > k$ we have $\mathcal{A}_n= \emptyset$.\\
Since $\mathcal{E}_n \neq \emptyset$ for all $n \in \mathbb{N}^*$, we can deduce that each $\mathcal{E}_n$ has a minimum by the well-ordering principle.
Let $m_n=\min_{m \in \mathcal{E}_n} \{m\}$ for all $n \in \mathbb{N}^*$ and let $$\{m_n\}_{n \in \mathbb{N}^*}=\{m_1, m_2, \ldots\}$$ be the sequence whose terms are the all the minima of each $\mathcal{E}_n$.\\
By definition, $$m_n = \prod_{j=1}^{n^e}p^*_j$$ where $p^*_1=2, p^*_2=3, p^*_3=5, \ldots$. The sequence $\{m_n\}$ is a positive term sequence and $$m_n = \prod_{j=1}^{n^2} p^*_j \geq 2^{n^e} \quad \forall n \in \mathbb{N}^*,$$ so $\{m_n\}_{n \in \mathbb{N}}$ diverges by the Comparison Test, i.e. $$\forall M \in \mathbb{R} \quad \exists k \in \mathbb{N} : m_n > M \quad \forall n>k$$ and this holds in particular for $M=10^{n^e}$, which gives $m_n > 10^{n^e}$ for any $n>k$. Since $m_n=\min_{m \in \mathcal{E}_n} \{m\}$, we have that $\mathcal{A}_n=\emptyset$ for any $n > k$ and for any fixed eponent $e$, which is the desired conclusion.

Mathematical Reflections 2017, Issue 3 - Problem U410

Problem:
Let $a,b,c$ be real numbers such that $a+b+c=5$. Prove that $$(a^2+3)(b^2+3)(c^2+3) \geq 192.$$

Proposed by Marius Stanean, Zalau, Romania

Solution:
Let $f(a,b,c)=(a^2+3)(b^2+3)(c^2+3)$ and $g(a,b,c)=a+b+c-5$. Consider the Lagrangian function
$$L(a,b,c,\lambda)=f(a,b,c)-\lambda g(a,b,c)=(a^2+3)(b^2+3)(c^2+3)-\lambda(a+b+c-5),$$
where $\lambda \in \mathbb{R}$. Observe that $f(a,b,c)$ is continuous on the compact set defined by $g(a,b,c)$, so by Weierstrass Theorem, there exists a global minimum. By the method of Lagrange Multipliers, a maximum or a minimum for $f(a,b,c)$ subject to the constraint $g(a,b,c)=0$ must be a stationary point of $L$. Therefore a maximum or a minimum satisfies
$$\renewcommand{\arraystretch}{2} \begin{array}{rcl} \dfrac{\partial L}{\partial a} & = & 0 \\ \dfrac{\partial L}{\partial b} & = & 0 \\  \dfrac{\partial L}{\partial c} & = & 0 \\ \dfrac{\partial L}{\partial \lambda} & = & 0, \end{array}$$ i.e.
$$\begin{array}{rcl} 2a(b^2+3)(c^2+3)+\lambda & = & 0 \\ 2b(a^2+3)(c^2+3)+\lambda & = & 0 \\ 2c(a^2+3)(b^2+3)+\lambda & = & 0 \\ a+b+c-5 & = & 0. \end{array}$$
Subtracting side by side the first two equations and simplifying, then doing the same for the second and the third equation and for the third and the first equation, we obtain
$$\begin{array}{rcl} (a-b)(3-ab) & = & 0 \\ (b-c)(3-bc) & = & 0 \\ (c-a)(3-ca) & = & 0 \\ a+b+c-5 & = & 0. \end{array}$$
So, up to symmetry, we get $$(a,b,c) \in \left\{\left(\dfrac{5}{3},\dfrac{5}{3},\dfrac{5}{3}\right),\left(\dfrac{3}{2},\dfrac{3}{2},2 \right),\left(1,1,3\right)\right\}.$$ Since $f\left(\dfrac{5}{3},\dfrac{5}{3},\dfrac{5}{3}\right)=\left(\dfrac{52}{9}\right)^3$, $f\left(\dfrac{3}{2},\dfrac{3}{2},2\right)=\dfrac{3087}{16}$, $f(1,1,3)=192$, we conclude that $192$ is the constrained minimum, i.e.
$$(a^2+3)(b^2+3)(c^2+3) \geq 192.$$

Mathematical Reflections 2017, Issue 3 - Problem U409

Problem:
Evaluate $$\lim_{x \to 0} \dfrac{2\sqrt{1+x}+2\sqrt{2^2+x}+\ldots+2\sqrt{n^2+x}-n(n+1)}{x}.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Observe that
$$\renewcommand{\arraystretch}{2} \begin{array}{rcl} \sqrt{1+x}&=&1+\dfrac{1}{2}\cdot x+o(x) \\ \sqrt{2^2+x}=2\sqrt{1+\dfrac{x}{2^2}}&=&2+\dfrac{1}{2}\cdot\dfrac{x}{2}+o(x) \\ \vdots & \vdots & \vdots \\ \sqrt{n^2+x}=n\sqrt{1+\dfrac{x}{n^2}}&=& n+\dfrac{1}{2}\cdot\dfrac{x}{n}+o(x). \end{array}$$
So, $$\begin{array}{lll} 2\sqrt{1+x}+2\sqrt{2^2+x}+\ldots+2\sqrt{n^2+x}&=&2(1+2+\ldots+n)+x\left(1+\dfrac{1}{2}+\ldots+\dfrac{1}{n}\right)+o(x)\\&=&n(n+1)+x\left(1+\dfrac{1}{2}+\ldots+\dfrac{1}{n}\right)+o(x), \end{array}$$ which gives
$$\lim_{x \to 0} \dfrac{2\sqrt{1+x}+2\sqrt{2^2+x}+\ldots+2\sqrt{n^2+x}-n(n+1)}{x}=1+\dfrac{1}{2}+\ldots+\dfrac{1}{n}.$$

Mathematical Reflections 2017, Issue 3 - Problem S414

Problem:
Prove that for any positive integers $a$ and $b$
$$(a^6-1)(b^6-1)+(3a^2b^2+1)(2ab-1)(ab+1)^2$$
is the product of at least four primes, not necessarily distinct.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Let $$N=(a^6-1)(b^6-1)+(3a^2b^2+1)(2ab-1)(ab+1)^2.$$ If $a=b=1$, then $N=16=2\cdot 2 \cdot 2 \cdot 2$. If $b=1$ and $a>1$, then $N=(3a^2+1)(2a-1)(a+1)^2$ is the product of four factors greater than $1$ and so is the product of at least four primes. Assume that $a \geq b>1$. Let $s=a+b$, $p=ab$ and $d=a-b$. Observe that $d^2=(a-b)^2=(a+b)^2-4ab=s^2-4p$. Clearly, $s \geq 4$, $p \geq 4$ and $d \geq 0$. We have
$$\begin{array}{lll} a^6+b^6&=&(a^2+b^2)(a^4+b^4)-(ab)^2(a^2+b^2)\\&=&(a^2+b^2)(a^4+b^4-a^2b^2)\\&=&(s^2-2p)(s^4+p^2-4s^2p). \end{array}$$
Hence, $$\begin{array}{lll} N&=&p^6-(s^2-2p)(s^4+p^2-4s^2p)+1+(3p^2+1)(2p-1)(p+1)^2\\&=&(p^2+4p-s^2)(p^4+2p^3+p^2s^2+p^2-2ps^2+s^4) \\ &=&(p^2-d^2)[(p^2+s^2-p)^2-d^2p^2]\\&=&(p-d)(p+d)(p^2+s^2-p-dp)(p^2+s^2-p+dp). \end{array}$$
Since $a \geq b>1$, then $p-d>1$ and $p+d>1$. Moreover, $p^2+s^2-p-dp=p(p-d)+s^2-p>p+s^2-p=s^2>1$ and $p^2+s^2-p+dp \geq p(p-1)+s^2>1$, so $N$ is the product of four factors greater than $1$ and so is the product of at least four primes.

Mathematical Reflections 2017, Issue 3 - Problem S412

Problem:
Let $a,b,c$ be positive real numbers such that
$$\dfrac{1}{\sqrt{1+a^3}}+\dfrac{1}{\sqrt{1+b^3}}+\dfrac{1}{\sqrt{1+c^3}} \leq 1.$$ Prove that $a^2+b^2+c^2 \geq 12$.

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Observe that for any positive real number $x$ we have
$$\dfrac{1}{1+\frac{1}{2}x^2} \leq \dfrac{1}{\sqrt{1+x^3}} \iff 1+x^3 \leq \left(1+\dfrac{1}{2}x^2\right)^2 \iff x^2(x-2)^2 \geq 0.$$
It follows that
$$\sum_{cyc} \dfrac{1}{1+\frac{1}{2}a^2} \leq  \sum_{cyc} \dfrac{1}{\sqrt{1+a^3}} \leq 1.$$
By Cauchy-Schwarz Inequality, we get
$$\begin{array}{lll} \displaystyle 1\cdot\left(3+\dfrac{1}{2}(a^2+b^2+c^2)\right) & \geq & \displaystyle \sum_{cyc} \dfrac{1}{1+\frac{1}{2}a^2}\sum_{cyc} \left(1+\dfrac{1}{2}a^2\right) \\ & \geq & 9, \end{array}$$
so $a^2+b^2+c^2 \geq 12$.

Mathematical Reflections 2017, Issue 3 - Problem S411

Problem:
Solve in real numbers the system of equations
$$\left\{\begin{array}{rcl} \sqrt{x}-\sqrt{y}&=&45 \\ \sqrt[3]{x-2017}-\sqrt[3]{y}&=&2. \end{array} \right.$$

Proposed by Adrian Andreescu, Dallas, Texas, USA

Solution:
Squaring both sides of the first equation, cubing both sides of the second equation and using $\sqrt[3]{x-2017}-\sqrt[3]{y}=2$, we obtain the system of equations
$$\begin{array}{rcl} x+y-2\sqrt{xy}&=&2025 \\ x-y-6\sqrt[3]{(x-2017)y}&=&2025. \end{array}$$
Subtracting side by side these two equations, we get
$$2\sqrt[3]{y}(\sqrt[3]{y^2}-\sqrt{x}\cdot \sqrt[6]{y}+3\sqrt[3]{x-2017})=0.$$
If $y=0$, we get $x=2025$. If $y \neq 0$, then $\sqrt[3]{y^2}-\sqrt{x}\cdot \sqrt[6]{y}+3\sqrt[3]{x-2017}=0$, i.e.
$$\sqrt[6]{y}(\sqrt{y}-\sqrt{x})+3\sqrt[3]{x-2017}=0 \iff -45\sqrt[6]{y}+3\sqrt[3]{x-2017}=0,$$ which gives $x=3375\sqrt{y}+2017$.
Substituting this equation into the first and the second equation of the system, we get
$$y=\left(\dfrac{16}{3285+223\sqrt{217}}\right)^2, \qquad y=\left(\dfrac{3285+223\sqrt{217}}{2}\right)^2.$$
So, $$(x,y)=(2025,0),$$ $$(x,y)=\left(2017+\dfrac{54000}{3285+223\sqrt{217}},\left(\dfrac{16}{3285+223\sqrt{217}}\right)^2\right),$$
$$(x,y)=\left(2017+\dfrac{3375}{2}(3285+223\sqrt{217}),\left(\dfrac{3285+223\sqrt{217}}{2}\right)^2\right).$$

Mathematical Reflections 2017, Issue 3 - Problem S409

Problem:
Solve in real numbers the equation $$2\sqrt{x-x^2}-\sqrt{1-x^2}+2\sqrt{x+x^2}=2x+1.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Clearly, $0 \leq x \leq 1$. Observe that
$$\renewcommand{\arraystretch}{2} \begin{array}{rcl} (\sqrt{x}+\sqrt{1-x})^2&=&1+2\sqrt{x-x^2} \\ \dfrac{1}{2}(\sqrt{1-x}+\sqrt{1+x})^2&=&1+\sqrt{1-x^2} \\ (\sqrt{x}-\sqrt{1+x})^2&=&2x+1-2\sqrt{x+x^2}. \end{array}$$
So, the given equation is equivalent to
$$(\sqrt{x}+\sqrt{1-x})^2-(\sqrt{x}-\sqrt{1+x})^2=\dfrac{1}{2}(\sqrt{1-x}+\sqrt{1+x})^2,$$
i.e. $$(\sqrt{1-x}+\sqrt{1+x})(2\sqrt{x}+\sqrt{1-x}-\sqrt{1+x})=\dfrac{1}{2}(\sqrt{1-x}+\sqrt{1+x})^2.$$
Since $\sqrt{1-x}+\sqrt{1+x}>0$, then the equation becomes $$2\sqrt{x}+\sqrt{1-x}-\sqrt{1+x}=\dfrac{1}{2}(\sqrt{1-x}+\sqrt{1+x}),$$ i.e. $$4\sqrt{x}=3\sqrt{1+x}-\sqrt{1-x}.$$ Squaring both sides, we get $$16x=9(1+x)+(1-x)-6\sqrt{1-x^2},$$ i.e. $$(1+x)+9(1-x)-6\sqrt{1-x^2}=0,$$ so $$\left(\sqrt{1+x}-3\sqrt{1-x}\right)^2=0.$$ It follows that $\sqrt{1+x}=3\sqrt{1-x}$, i.e. $1+x=9(1-x)$, which gives $x=\dfrac{4}{5}$.

Mathematical Reflections 2017, Issue 3 - Problem J414

Problem:
Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that
$$\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2} \geq a^2+b^2+c^2.$$

Proposed by Dragoljub Milosevici, Gornji Milanovac, Serbia

Solution:
By the AM-GM Inequality, we have
$$\renewcommand{\arraystretch}{2} \begin{array}{lll} \dfrac{a^3}{b^2}+ab^2 & \geq & 2a^2 \\ \dfrac{b^3}{c^2}+bc^2 & \geq & 2b^2 \\ \dfrac{c^3}{a^2}+ca^2 & \geq & 2c^2 \end{array}$$
So, $$\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2} \geq 2(a^2+b^2+c^2)-(ab^2+bc^2+ca^2).$$
We want to prove that $2(a^2+b^2+c^2)-(ab^2+bc^2+ca^2) \geq a^2+b^2+c^2$, i.e.

$$a^2+b^2+c^2 \geq ab^2+bc^2+ca^2   \qquad (1)$$

By the AM-GM Inequality, we have $$\begin{array}{lll} a^3+ac^2 & \geq & 2ca^2 \\ b^3+ba^2 & \geq & 2ab^2 \\ c^3+cb^2 & \geq & 2bc^2, \end{array}$$
so $$a^3+b^3+c^3+ac^2+ba^2+cb^2 \geq 2(ab^2+bc^2+ca^2),$$ i.e. $$(a+b+c)(a^2+b^2+c^2) \geq 3(ab^2+bc^2+ca^2).$$ Since $a+b+c=3$, we get inequality (1) and the conclusion follows. Equality holds if and only if $a=b=c=1$.

Mathematical Reflections 2017, Issue 3 - Problem J413

Problem:
Solve in integers the system of equations $$\begin{array}{lll} x^2y+y^2z+z^2x-3xyz&=&23 \\ xy^2+yz^2+zx^2-3xyz&=&25. \end{array}$$

Proposed by Adrian Andreescu, Dallas, USA

Solution:
Subtracting the first equation to the second equation, we have
$$xy(y-x)+yz(z-y)+zx(x-z)=2,$$ i.e. $$(x-y)(y-z)(z-x)=2.$$ Since $(x-y)+(y-z)+(z-x)=0$ and $2+(-1)+(-1)=(-2)+1+1=0$, then
$$\begin{array}{lll} x-y&=&-1 \\ y-z&=&-1 \\ z-x&=&2, \end{array} \qquad \begin{array}{lll} x-y&=&-1 \\ y-z&=&2 \\ z-x&=&-1, \end{array} \qquad \begin{array}{lll} x-y&=&2 \\ y-z&=&-1 \\ z-x&=&-1, \end{array}$$
$$\begin{array}{lll} x-y&=&1 \\ y-z&=&1 \\ z-x&=&-2, \end{array} \qquad \begin{array}{lll} x-y&=&1 \\ y-z&=&-2 \\ z-x&=&1, \end{array} \qquad \begin{array}{lll} x-y&=&-2 \\ y-z&=&1 \\ z-x&=&1. \end{array}$$
Hence, $$\begin{array}{lll} (x,y,z) & \in & \{(n,n+1,n+2),(n,n+1,n-1),(n,n-2,n-1),\\& & (n,n-1,n-2),(n,n-1,n+1),(n,n+2,n+1)\}. \end{array}$$
Substituting each triple into the first equation of the system and with an easy check to the second equation, we get
$$(x,y,z) \in \{(7,8,9),(8,9,7),(9,7,8)\}.$$

Mathematical Reflections 2017, Issue 3 - Problem J412

Problem:
Let $a \geq b \geq c$ be positive real numbers. Prove that
$$(a-b+c)\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}+\dfrac{1}{c+a}\right) \leq \dfrac{1}{2}.$$

Proposed by An Zhenping, Xianyang Normal University, China

Solution:
If $a \geq b \geq c$ are positive real numbers, then $a=c+x+y$ and $b=c+y$, where $x,y \geq 0$. The given inequality is equivalent to
$$(x-c)\left(\dfrac{1}{2c+x+2y}-\dfrac{1}{2c+y}+\dfrac{1}{2c+x+y}\right) \leq \dfrac{1}{2}.$$
This inequality is equivalent to
$$\dfrac{16 c^3 + 24 c^2 y + 12 c y^2 + 2 x^3 + 3 x^2 y + x y^2 + 2 y^3}{(2c+x+2y)(2c+y)(2c+x+y)} \geq 0,$$
which is clearly true.

Mathematical Reflections 2017, Issue 3 - Problem J411

Problem:
Find all primes $p$ and $q$ such that $$\dfrac{p^3-2017}{q^3-345}=q^3.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
The given equation is equivalent to $$p^3-2017=q^6-345q^3.$$ Reducing modulo $7$, we get $$p^3-1 \equiv q^6-2q^3 \pmod{7} \iff p^3 \equiv (q^3-1)^2 \pmod{7}.$$
Since $q^3 \equiv 0,1,6 \pmod{7}$, then $(q^3-1)^2 \equiv 0,1,4 \pmod{7}$. Since $p^3 \equiv 0,1,6 \pmod{7}$, we get $(q^3-1)^2 \equiv 0,1 \pmod{7}$. If $(q^3-1)^2 \equiv 0 \pmod{7}$, then $p^3 \equiv 0 \pmod{7}$, which gives $p=7$. But the equation $q^6-345q^3+1674=0$ gives no integer solutions. If $(q^3-1)^2 \equiv 1 \pmod{7}$, then $q^3 \equiv 0 \pmod{7}$, which gives $q=7$. So, $p^3-2017=-686$, i.e. $p^3=1331$, which gives $p=11$. So, $p=11$ and $q=7$.

Mathematical Reflections 2017, Issue 3 - Problem J410

Problem:
Let $a,b,c,d$ be real numbers such that $a^2 \leq 2b$ and $c^2<2bd$. Prove that $$x^4+ax^3+bx^2+cx+d>0$$ for all $x \in \mathbb{R}$.

Proposed by Dorin Andrica, Babes-Bolyai University, Cluj-Napoca, Romania

Solution:
Since $a^2 \leq 2b$ and $c^2<2bd$, we have $b \geq 0$ and $d>0$. Moreover,
$$x^4+ax^3+bx^2+cx+d=\left(x^2+\dfrac{a}{2}x\right)^2+\left(b-\dfrac{a^2}{4}\right)x^2+cx+d.$$ Clearly, $\left(x^2+\dfrac{a}{2}x\right)^2 \geq 0$. Since $b-\dfrac{a^2}{4}>0$ and the discriminant of $\left(b-\dfrac{a^2}{4}\right)x^2+cx+d$ is $$c^2-4\left(b-\dfrac{a^2}{4}\right)d=c^2-4bd+a^2d<-2bd+a^2d=d(a^2-2b) \leq 0,$$ then $$\left(b-\dfrac{a^2}{4}\right)x^2+cx+d>0$$ for all $x \in \mathbb{R}$, which gives $x^4+ax^3+bx^2+cx+d>0$ for all $x \in \mathbb{R}$.

Mathematical Reflections 2017, Issue 3 - Problem J409

Problem:Solve the equation $$\log(1-2^x+5^x-20^x+50^x)=2x.$$

Proposed by Adrian Andreescu, Dallas, Texas, USA

Solution:
The given equation is equivalent to $$1-2^x+5^x-20^x+50^x=10^{2x},$$ i.e. $$1-2^x+10^x-20^x+5^x-10^x+50^x-100^x=0,$$ which gives
$$(1-2^x)(1+5^x)(1+10^x)=0.$$ It follows that $1-2^x=0$, so $x=0$.

Mathematical Reflections 2017, Issue 2 - Problem O406

Problem:
Solve in prime numbers the equation $$x^3-y^3-z^3+w^3+\dfrac{yz}{2}(2xw+1)^2=2017.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Since $x^3-y^3-z^3+w^3$ and $2017$ are integers, then also $\dfrac{yz}{2}(2xw+1)^2$ must be an integer. Since $(2xw+1)^2$ is odd, then $2 \ | \ yz$. Since $y$ and $z$ are primes, then $y=2$ or $z=2$. By symmetry, we can assume $y=2$. We have
$$x^3-z^3+w^3+z(2xw+1)^2=2025.$$ Since $2025$ is odd, then the LHS must be odd and it's easy to see that $x,z,w$ cannot be all odd. Moreover, the given equation can be written as $$x^3+w^3+z(2xw-z+1)(2xw+z+1)=2025.$$ Since $z(2xw-z+1)(2xw+z+1)$ is always even, then $x^3+w^3$ must be odd, which implies that at least one between $x$ and $w$ is even, i.e. $x=2$ or $w=2$. By symmetry, we can assume $x=2$. We have $$w^3+z(4w-z+1)(4w+z+1)=2017.$$ If $4w-z+1 \geq 0$, then $w^3 \leq 2017$, so $w \leq 11$. An easy check gives $w=7$ and $z=2$.
So, $(x,y,z,w) \in \{(2,2,2,7),(7,2,2,2)\}$.
The case $4w-z+1 < 0$ remains open.

Mathematical Reflections 2017, Issue 2 - Problem U407

Problem:
Prove that for every $\varepsilon>0$ $$\int_2^{2+\varepsilon} e^{2x-x^2} \ dx<\dfrac{\varepsilon}{1+\varepsilon}.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Observe that for all real numbers $t \geq 1$ we have $t^2 \leq e^{t^2-1}$. Indeed, put $f(t)=e^{t^2-1}-t^2$. Then, $f(1)=0$ and $f'(t)=2t(e^{t^2-1}-1) \geq 0$ for all $t \geq 1$. Now, put $t=x-1$. We have $(x-1)^2 \leq e^{x^2-2x}$ for all $x \geq 2$, i.e. $$e^{2x-x^2} \leq \dfrac{1}{(x-1)^2}.$$
So, $$\int_2^{2+\varepsilon} e^{2x-x^2} \ dx \leq \int_2^{2+\varepsilon} \dfrac{dx}{(x-1)^2}=\left[-\dfrac{1}{x-1}\right]_2^{2+\varepsilon}=\dfrac{\varepsilon}{1+\varepsilon}.$$

Mathematical Reflections 2017, Issue 2 - Problem U406

Problem:
Evaluate $$\lim_{x \to 0} \dfrac{\cos(n+1)x\cdot \sin nx-n\sin x}{x^3}.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Observe that $$\renewcommand{\arraystretch}{2} \begin{array}{rcl} \cos(n+1)x&=&1-\dfrac{1}{2}(n+1)^2x^2+o(x^3) \\  \sin nx&=&nx-\dfrac{1}{6}n^3x^3+o(x^3) \\ n\sin x&=&nx-\dfrac{1}{6}nx^3+o(x^3). \end{array}$$
So, $$\dfrac{\left(1-\frac{1}{2}(n+1)^2x^2+o(x^3)\right)\left(nx-\frac{1}{6}n^3x^3+o(x^3)\right)-\left(nx-\frac{1}{6}nx^3+o(x^3)\right)}{x^3}=-\dfrac{\frac{x^3}{3}n(n+1)(2n+1)+o(x^3)}{x^3}.$$ So, $$\lim_{x \to 0} \dfrac{\cos(n+1)x\cdot \sin nx-n\sin x}{x^3}=-\dfrac{n(n+1)(2n+1)}{3}.$$

Mathematical Reflections 2017, Issue 2 - Problem U403

Problem:
Find all cubic polynomials $P(x) \in \mathbb{R}[x]$ such that $$P\left(1-\dfrac{x(3x+1)}{2}\right)-P^2(x)+P\left(\dfrac{x(3x-1)}{2}-1\right)=1 \qquad \forall x \in \mathbb{R}.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Let $P(x)=ax^3+bx^2+cx+d$ be a third-degree polynomial with $a,b,c,d \in \mathbb{R}$, $a \neq 0$, such that
$$P\left(1-\dfrac{x(3x+1)}{2}\right)-P^2(x)+P\left(\dfrac{x(3x-1)}{2}-1\right)=1 \qquad \forall x \in \mathbb{R}.$$
Setting $x=-1,x=0,x=1$ into the given relation, we obtain
$$\begin{array}{rcl} P(0)-P^2(-1)+P(1)&=&1 \\ P(1)-P^2(0)+P(-1)&=&1 \\ P(-1)-P^2(1)+P(0)&=&1. \end{array}$$
Adding these three equations side by side, we get $$(P(-1)-1)^2+(P(0)-1)^2+(P(1)-1)^2=0,$$ so $P(-1)=P(0)=P(1)=1$. It follows that
$$\begin{array}{rcl} d&=&1 \\ a+b+c+d&=&1 \\ -a+b-c+d&=&1. \end{array}$$ We obtain $d=1,b=0$ and $c=-a$. So, $P(x)=ax^3-ax+1$.
Setting $x=2$ into the given relation, we have $$P(-6)-P^2(2)+P(4)=1,$$ which gives $$-36a^2-162a+1=1.$$ Likewise, setting $x=-2$ into the given relation, we have $$P(-4)-P^2(-2)+P(6)=1,$$ which gives $$-36a^2+162a+1=1.$$ Subtracting, we get $264a=0$, i.e. $a=0$. So, we obtain $P(x)=1$.

Mathematical Reflections 2017, Issue 2 - Problem S407

Problem:
Let $f(x)=x^3+x^2-1$. Prove that for any positive real numbers $a, b, c, d$ satisfying $$a+b+c+d>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d},$$ at least one of the numbers $af(b)$, $bf(c)$, $cf(d)$, $df(a)$ is different from $1$.

Proposed by Adrian Andreescu, Dallas, Texas, USA

Solution:
Assume by contradiction that $af(b)=bf(c)=cf(d)=df(a)=1$. The given inequality becomes
$$\begin{equation}\label{first-ineq} a+b+c+d>f(a)+f(b)+f(c)+f(d). \end{equation}$$
Observe that $f(x)$ is increasing on $(0,+\infty)$ and $f(x)<x$ if and only if $x<1$. Moreover, $f$ is convex on $(0,+\infty)$, so by inequality $\eqref{first-ineq}$ and Jensen's Inequality, we have $$\dfrac{a+b+c+d}{4}>\dfrac{f(a)+f(b)+f(c)+f(d)}{4} \geq f\left(\dfrac{a+b+c+d}{4}\right).$$ It follows that $\dfrac{a+b+c+d}{4}<1$, i.e. $a+b+c+d<4$ and from $\eqref{first-ineq}$, we get
$$f(a)+f(b)+f(c)+f(d)<4.$$ On the other hand, by the AM-GM Inequality, we have $$\begin{array}{lll} af(a)+bf(b)+cf(c)+df(d) & \geq & 4 \\ bf(a)+cf(b)+df(c)+af(d) & \geq & 4 \\ cf(a)+df(b)+af(c)+bf(d) & \geq & 4 \\ df(a)+af(b)+bf(c)+cf(d) & \geq & 4. \end{array}$$ Adding these four inequalities, we get $(a+b+c+d)(f(a)+f(b)+f(c)+f(d))\geq 16$, i.e. $$f(a)+f(b)+f(c)+f(d) \geq \dfrac{16}{a+b+c+d} > 4,$$ contradiction.
The conclusion follows.

Mathematical Reflections 2017, Issue 2 - Problem S403

Problem:
Find all primes $p$ and $q$ such that $$\dfrac{2^{p^2-q^2}-1}{pq}$$ is a product of two primes.

Proposed by Adrian Andreescu, Dallas, Texas, USA

Solution:
Since $2^{p^2-q^2}-1$ is odd and $$\dfrac{2^{p^2-q^2}-1}{pq}$$ is an integer, then $p$ and $q$ are odd primes and clearly $p>q$. So, $p^2-q^2 \equiv 0 \pmod{8}$. If $q>3$, then $p^2-q^2 \equiv 0 \pmod{3}$, so $p^2-q^2$ is divisible by $24$. Then
$$2^{p^2-q^2}-1=2^{24k}-1,$$ where $k \in \mathbb{N}^*$. Since $2^{24k}-1$ is divisible by $2^{24}-1=16777215=3^2\cdot 5\cdot 7 \cdot 13 \cdot 17 \cdot 241$, then $\dfrac{2^{p^2-q^2}-1}{pq}$ cannot be the product of two primes. So, $q=3$ and
$$\dfrac{2^{p^2-q^2}-1}{pq}=\dfrac{2^{p^2-9}-1}{3p}.$$ Since $p^2-9$ is divisible by $8$, then $p^2-9=8k$ for some $k \in \mathbb{N}^*$, so $$\dfrac{2^{p^2-9}-1}{3p}=\dfrac{2^{8k}-1}{3p}=\dfrac{(2^k-1)(2^k+1)(2^{2k}+1)(2^{4k}+1)}{3p}.$$ An easy check shows that it must be $k \geq 2$. If $k$ is even, then $3 \ | \ (2^k-1)$ and if $k>2$ we have $2^k-1=3n$ for some $n \in \mathbb{N}$ and $n>1$. But then
$$\dfrac{n(2^k+1)(2^{2k}+1)(2^{4k}+1)}{p}$$ is the product of at least three primes, contradiction. So, $k=2$ and we get $p=5$, which satisfies the condition. If $k$ is odd, then $3 \ | \ (2^k+1)$ and since $k \geq 3$, then $2^k+1=3n$ for some $n \in \mathbb{N}$ and $n>1$. But then, we have that $$\dfrac{n(2^k-1)(2^{2k}+1)(2^{4k}+1)}{p}$$ is the product of at least three primes, contradiction.
In conclusion, $(p,q)=(5,3)$.

Mathematical Reflections 2017, Issue 2 - Problem J408

Problem:
Let $a$ and $b$ be nonnegative real numbers such that $a+b=1$. Prove that $$\dfrac{289}{256} \leq (1+a^4)(1+b^4) \leq 2.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
By the Cauchy-Schwarz Inequality, we have $$(1+a^4)(1+b^4) \geq (1+a^2b^2)^2=(1+a^2(1-a)^2)^2 \geq \left(1+\dfrac{1}{16}\right)^2=\dfrac{289}{256}$$ and the minimum is attained if and only if $a=b=\dfrac{1}{2}$. By the AM-GM Inequality, we have
$$(1+a^4)(1+b^4) \leq \dfrac{(1+a^4)^2+(1+b^4)^2}{2}=\dfrac{(1+a^4)^2+(1+(1-a)^4)^2}{2}.$$
Using the fact $$\begin{array}{rcll} a^4 & \leq & (1-a)^4, & a \in [0,1/2] \\ (1-a)^4 & \leq & a^4, & a \in [1/2,1], \end{array}$$ then if $a \in [0,1/2]$, we have
$$\dfrac{(1+a^4)^2+(1+(1-a)^4)^2}{2} \leq (1+(1-a)^4)^2 \leq 2$$
and if $a \in [1/2,1]$, we have
$$\dfrac{(1+a^4)^2+(1+(1-a)^4)^2}{2} \leq (1+a^4)^2 \leq 2.$$
The conclusion follows.

Mathematical Reflections 2017, Issue 2 - Problem J407

Problem:
Solve in positive real numbers the equation $$\sqrt{x^4-4x}+\dfrac{1}{x^2}=1.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

First solution:
The equation has a real solution if and only if $x \leq -1$ or $x \geq \sqrt[3]{4}$. We have $$x^4-4x=\left(1-\dfrac{1}{x^2}\right)^2 \iff x^4-4x+\dfrac{4}{x^2}=\left(1-\dfrac{1}{x^2}\right)^2+\dfrac{4}{x^2},$$ i.e.
$$\left(x^2-\dfrac{2}{x}\right)^2=\left(1+\dfrac{1}{x}\right)^2$$
So, $$x^2-\dfrac{2}{x}=\pm \left(1+\dfrac{1}{x^2}\right).$$
Observe that $x^2-\dfrac{2}{x}=-1-\dfrac{1}{x^2}$ gives $x^4+(x-1)^2=0$, which has no real solutions. So,
$$x^2-\dfrac{2}{x}=1+\dfrac{1}{x^2} \iff x(x^3-1)=x^2+x+1 \iff x(x-1)=1.$$ So, we get $x^2-x-1=0$, which yields $x=\dfrac{1+\sqrt{5}}{2}$.

Second solution:
The given equation is equivalent to $$x\sqrt{x^4-4x}=x-\dfrac{1}{x}.$$ Squaring both sides, we get $x^6-4x^3=x^2-2+\dfrac{1}{x^2}$. Hence, $$(x^3-2)^2=\left(x+\dfrac{1}{x}\right)^2.$$ Since $x^3>4$, we have $x^3-2=x+\dfrac{1}{x}$. So, $x^4-2x=x^2+1$, which gives $x^4=(x+1)^2$, i.e. $x^2=x+1$. So, $x=\dfrac{1+\sqrt{5}}{2}$.

Mathematical Reflections 2017, Issue 2 - Problem J406

Problem:
Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that $$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Let $f(x)=x\sqrt{x+3}$. Observe that $$f''(x)=\dfrac{3(x+4)}{4\sqrt{(x+3)^3}}>0 \qquad \forall x>0.$$ Therefore, by Jensen's Inequality, we have $$2=f(1)=f\left(\dfrac{a+b+c}{3}\right) \leq \dfrac{f(a)+f(b)+f(c)}{3}=\dfrac{a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3}}{3},$$ i.e. $$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6.$$

Mathematical Reflections 2017, Issue 1 - Problem O397

Problem:
Solve in integers the equation: $$(x^3-1)(y^3-1)=3(x^2y^2+2).$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
The given equation can be written as $$x^3y^3-(x^3+y^3)-3x^2y^2=5.$$ Let $s=x+y$ and $t=xy$. Observe that $x^3+y^3=(x+y)^3-3xy(x+y)=s^3-3st$, so the given equation becomes $$(t^3-s^3)-3t(t-s)=5,$$ i.e. $$(t-s)(t^2+ts+s^2-3t)=5.$$
We obtain the four systems of equations: $$\begin{array}{rcl} t-s&=&\pm 1 \\ t^2+ts+s^2-3t&=& \pm 5, \end{array} \qquad \begin{array}{rcl} t-s&=&\pm 5 \\ t^2+ts+s^2-3t&=& \pm 1 \end{array}$$
If $t-s=\pm 1$, then $t^2-2ts+s^2=1$ and subtracting this equation to the second equation, we get $3ts-3t=4$ or $3ts-3t=-6$. The first equation is impossible, the second gives $t(s-1)=-2$. So, $(s,t) \in \{(-1,1),(0,2),(2,-2),(3,-1)\}$. It's easy to see that none of these pairs satisfies $t-s=\pm 1$, so there are no solutions in this case. If $t-s=\pm 5$, then $t^2-2ts+s^2=25$ and subtracting this equation to the second equation, we get $3ts-3t=-24$ or $3ts-3t=-26$. The second equation is impossible, the first gives $t(s-1)=-8$. So, $(s,t) \in \{(-7,1),(-3,2),(-1,4),(0,8),(2,-8),(3,-4),(5,-2),(9,-1)\}$. As $t-s=\pm 5$, we obtain $(s,t) \in \{(-3,2),(-1,4)\}$. Since $s$ and $t$ also satisfy the condition $s^2-4t=n^2$ for some $n \in \mathbb{Z}$, we obtain $(s,t)=(-3,2)$. So, $x+y=-3$ and $xy=2$, which gives $(x,y) \in \{(-1,-2),(-2,-1)\}$.

Mathematical Reflections 2017, Issue 1 - Problem U402

Problem:
Let $n$ be a positive integer and let $P(x)$ be a polynomial of degree at most $n$ such that $|P(x)| \leq x+1$ for all $x \in [0,n]$. Prove that
$$|P(n+1)|+|P(-1)| \leq (n+2)(2^{n+1}-1)$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
In order to prove the inequality, we deduce upper bounds for $|P(-1)|$ and $|P(n+1)|$.
Let $a_0,a_1,\ldots,a_n$ be $n+1$ distinct real numbers. Let $$L_i(x)=\prod_{\begin{smallmatrix} j=0\\ j\neq i\end{smallmatrix}}^{n} \dfrac{x-a_j}{a_i-a_j}$$ be the $n$-th degree Lagrange base polynomials for $i=0,1,\ldots,n$. We first prove that $\{L_0(x),\ldots,L_n(x)\}$ is a basis for $\mathbb{R}_n[x]$. Since $\dim \mathbb{R}_n[x]=n+1$, we only have to prove that $\{L_0(x),\ldots,L_n(x)\}$ are linearly independent. Let $(\lambda_0,\ldots,\lambda_{n+1}) \in \mathbb{R}^{n+1}$ such that $$\lambda_0L_0(x)+\ldots+\lambda_n L_n(x)=0$$ For each $i \in \{0,1,\ldots,n\}$, if $x=a_i$ we have $$0=\sum_{j=0}^n \lambda_j L_j(a_i)=\sum_{j=0}^n \lambda_j \delta_{ji}=\lambda_i,$$ so $\lambda_i=0$ for any $i \in \{0,1,\ldots,n\}$ and this shows that $\{L_0(x),\ldots,L_n(x)\}$ is a basis for $\mathbb{R}_n[x]$. So, if $P \in \mathbb{R}_n[x]$, there exists $(\lambda_0,\ldots,\lambda_n) \in \mathbb{R}^{n+1}$ such that $P(x)=\sum_{j=0}^n \lambda_jL_j(x)$. If $x=a_i$, then $$P(a_i)=\sum_{j=0}^n \lambda_j L_j(a_i)=\sum_{j=0}^n \lambda_j \delta_{ji}=\lambda_i,$$ so we can write $$P(x)=\sum_{j=0}^n P(a_j)L_j(x).$$ If $(a_0,a_1,\ldots,a_n)=(0,1,\ldots,n)$, we have $$P(x)=\sum_{j=0}^n P(j)L(x)$$
Now, we have $$\renewcommand{\arraystretch}{2} \begin{array}{lcl} \displaystyle |L_i(-1)|=\left| \prod_{\begin{smallmatrix} j=0\\ j\neq i\end{smallmatrix}}^{n} \dfrac{-1-j}{i-j} \right|&=& \displaystyle \prod_{j=0}^{i-1} \dfrac{j+1}{i-j} \prod_{j=i+1}^n \dfrac{j+1}{j-i}\\&=& \displaystyle \dfrac{i!}{i!}\cdot \dfrac{(n+1)!}{(i+1)!(n-i)!}\\&=& \displaystyle {n+1 \choose i+1} \end{array}$$
and
$$\renewcommand{\arraystretch}{2} \begin{array}{lcl} \displaystyle |L_i(n+1)|=\left| \prod_{\begin{smallmatrix} j=0\\ j\neq i\end{smallmatrix}}^{n} \dfrac{n+1-j}{i-j} \right|&=& \displaystyle \prod_{j=0}^{i-1} \dfrac{n+1-j}{i-j} \prod_{j=i+1}^n \dfrac{n+1-j}{j-i}\\&=& \displaystyle \dfrac{(n+1)!}{i!(n+1-i)!}\cdot \dfrac{(n-i)!}{(n-i)!}\\&=& \displaystyle {n+1 \choose i}. \end{array}$$
Hence, $$\renewcommand{\arraystretch}{2} \begin{array}{lcl} \displaystyle |P(-1)|=\left| \sum_{i=0}^n P(i)L_i(-1) \right| & \leq & \displaystyle \sum_{i=0}^n |P(i)L_i(-1)| \\ & \leq & \displaystyle \sum_{i=0}^n (i+1) {n+1 \choose i+1}\\&=& \displaystyle (n+1) \sum_{i=0}^n {n \choose i}\\&=&(n+1)2^n \end{array}$$ and
$$\renewcommand{\arraystretch}{2} \begin{array}{lcl} \displaystyle |P(n+1)|=\left| \sum_{i=0}^n P(i)L_i(n+1) \right| & \leq & \displaystyle \sum_{i=0}^n |P(i)L_i(n+1)| \\ & \leq & \displaystyle \sum_{i=0}^n (i+1) {n+1 \choose i} \\ &=& \displaystyle \sum_{i=0}^n i{n+1 \choose i}+\sum_{i=0}^n {n+1 \choose i} \\ &=& \displaystyle (n+1) \sum_{i=1}^n {n \choose i-1}+ \sum_{i=0}^n {n+1 \choose i} \\&=&(n+1)(2^n-1)+(2^{n+1}-1). \end{array}$$
Adding the last two inequalities, we get the desired inequality.

Mathematical Reflections 2017, Issue 1 - Problem U401

Problem:
Let $P$ be a polynomial of degree $n$ such that $P(k)=\dfrac{1}{k^2}$ for all $k=1,2,\ldots,n+1$. Determine $P(n+2)$.

Proposed by Dorin Andrica, Babe\c{s}-Bolyai University, Cluj-Napoca, Romania

Solution:
There exists a unique interpolating polynomial $P$ of degree $n$ such that $P(k)=\dfrac{1}{k^2}$ for all $k=1,2,\ldots,n+1$ and this is
$$P(x)=\sum_{k=1}^{n+1} \left(\prod_{\stackrel{1\leq j\leq n+1}{j\neq k}}\frac{x-j}{k-j}\right)\dfrac{1}{k^2}.$$
Observe that $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle \prod_{\stackrel{1\leq j\leq n+1}{j\neq k}}\frac{n+2-j}{k-j}&=&\displaystyle\prod_{j=1}^{k-1} \left(\dfrac{n+2-j}{k-j}\right)\prod_{j=k+1}^{n+1} \left(\dfrac{n+2-j}{k-j}\right)\\&=& \displaystyle \dfrac{(n+1)!}{(n-k+2)!(k-1)!}\cdot\dfrac{(n-k+1)!}{(-1)^{n-k+1}(n-k+1)!}\\&=& \displaystyle (-1)^{n-k+1}{n+1 \choose k-1}. \end{array}$$
So, $$P(n+2)=\sum_{k=1}^{n+1} (-1)^{n-k+1}{n+1 \choose k-1}\dfrac{1}{k^2}.$$

Mathematical Reflections 2017, Issue 1 - Problem U397

Problem:
Let $T_n$ be the $n$-th triangular number. Evaluate $$\sum_{n \geq 1} \dfrac{1}{(8T_n-3)(8T_{n+1}-3)}$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Let $t_n=\dfrac{1}{(8T_n-3)(8T_{n+1}-3)}$.
Observe that $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \dfrac{1}{t_n}=(8T_n-3)(8T_{n+1}-3)&=&\left(8\dfrac{n(n+1)}{2}-3\right)\left(8\dfrac{(n+1)(n+2)}{2}-3\right)\\&=&(4n^2+4n-3)(4n^2+12n+5)\\&=&(2n-1)(2n+3)(2n+1)(2n+5). \end{array}$$
We get $$\renewcommand{\arraystretch}{2} \begin{array}{lll} t_n&=&\dfrac{1}{(2n-1)(2n+3)(2n+1)(2n+5)}\\&=&\dfrac{1}{8}\dfrac{1}{(2n-1)(2n+5)}-\dfrac{1}{8}\dfrac{1}{(2n+1)(2n+3)}\\&=&\dfrac{1}{48}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+5}\right)-\dfrac{1}{16}\left(\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\right)\\&=&\dfrac{1}{48}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)+\dfrac{1}{48}\left(\dfrac{1}{2n+3}-\dfrac{1}{2n+5}\right)-\dfrac{1}{24}\left(\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\right). \end{array}$$
So, $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle \sum_{n \geq 1} t_n&=& \displaystyle \dfrac{1}{48}\sum_{n=1}^\infty \left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)+\dfrac{1}{48}\sum_{n=1}^\infty \left(\dfrac{1}{2n+3}-\dfrac{1}{2n+5}\right)-\dfrac{1}{24}\sum_{n=1}^\infty \left(\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\right) \\ &=&\dfrac{1}{48}+\dfrac{1}{48}\cdot\dfrac{1}{5}-\dfrac{1}{24}\cdot\dfrac{1}{3} \\ &=& \dfrac{1}{90}. \end{array}$$

Mathematical Reflections 2017, Issue 1 - Problem S402

Problem:
Prove that $$\sum_{k=1}^{31} \dfrac{k}{(k-1)^{4/5}+k^{4/5}+(k+1)^{4/5}}<\dfrac{3}{2}+\sum_{k=1}^{31} (k-1)^{1/5}.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Observe that $$\dfrac{k}{(k-1)^{4/5}+k^{4/5}+(k+1)^{4/5}}<\dfrac{k}{3(k-1)^{4/5}},$$ so
$$\begin{array}{lll} \displaystyle \sum_{k=1}^{31} \left(\dfrac{k}{(k-1)^{4/5}+k^{4/5}+(k+1)^{4/5}}-(k-1)^{1/5}\right)&<& \displaystyle \dfrac{1}{1+2^{4/5}}-\dfrac{1}{3}\sum_{k=2}^{31} \dfrac{2k-3}{(k-1)^{4/5}} \\ &<& \displaystyle \dfrac{1}{1+2^{4/5}}-\dfrac{1}{3}\sum_{k=2}^{31} \dfrac{2k-3}{k-1}\\&<& \displaystyle \dfrac{1}{1+2^{4/5}}-\dfrac{1}{3}\left(1+\dfrac{3}{2}\right)<0. \end{array}$$
So, $$\sum_{k=1}^{31} \left(\dfrac{k}{(k-1)^{4/5}+k^{4/5}+(k+1)^{4/5}}-(k-1)^{1/5}\right)<0<\dfrac{3}{2}.$$

Mathematical Reflections 2017, Issue 1 - Problem S400

Problem:
Find all $n$ for which $(n-4)!+\dfrac{1}{36n}(n+3)!$ is a perfect square.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Clearly, $n \geq 4$. We have $$\begin{array}{lll} (n-4)!+\dfrac{1}{36n}(n+3)!&=&(n-4)!\left(1+\dfrac{1}{36n}(n-3)(n-2)(n-1)n(n+1)(n+2)(n+3)\right)\\&=&(n-4)!\left(1+\dfrac{1}{36}(n^2-9)(n^2-4)(n^2-1)\right)\\&=&(n-4)!\left(\dfrac{n^6-14n^4+49n^2}{36}\right)\\&=&(n-4)!\left(\dfrac{n(n^2-7)}{6}\right)^2. \end{array}$$
Observe that $n(n^2-7)=n^3-7n \equiv n^3-n \equiv 0 \pmod{6}$, so $\dfrac{n(n^2-7)}{6}$ is an integer. It follows that $(n-4)!\left(\dfrac{n(n^2-7)}{6}\right)^2$ is a perfect square if and only if $(n-4)!$ is a perfect square. If $n=4$ or $n=5$, then $(n-4)!=1$, which is a perfect square. Let $n>5$ and let $p$ be the greatest prime that divides $(n-4)!$. By Bertrand's Postulate, there exists a prime $q$ such that $p<q<2p$. If $2p \leq n-4$, then $q<n-4$, which gives $q \ | \ (n-4)!$, contradiction. So, $n-4<2p$, which means that $p \ | \ (n-4)!$ and $p^2 \nmid (n-4)!$. So, $(n-4)!$ is not a perfect square if $n>5$. Therefore, $n \in \{4,5\}$.

Mathematical Reflections 2017, Issue 1 - Problem S397

Problem:
Let $a,b,c$ be positive real numbers. Prove that $$\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}+\dfrac{3(ab+bc+ca)}{2(a+b+c)} \geq a+b+c.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
The given inequality is equivalent to $$(a+b+c)\left(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\right)+\dfrac{3}{2}(ab+bc+ca) \geq (a+b+c)^2,$$ i.e.

$$\dfrac{a^2c}{a+b}+\dfrac{b^2a}{b+c}+\dfrac{c^2b}{c+a} \geq \dfrac{1}{2}(ab+bc+ca) \qquad (1)$$

By the AM-GM Inequality, we have $$\dfrac{2a^2c}{a+b}+\dfrac{c(a+b)}{2} \geq 2ca,$$ $$\dfrac{2b^2a}{b+c}+\dfrac{a(b+c)}{2} \geq 2ab,$$ $$\dfrac{2c^2b}{c+a}+\dfrac{b(c+a)}{2} \geq 2bc.$$ Adding these three inequalities, we get inequality (1). The equality holds if and only if $a=b=c$.

Mathematical Reflections 2017, Issue 1 - Problem J401

Problem:
Find all integers $n$ for which $n^2+2^n$ is a perfect square.

Proposed by Adrian Andreescu, Dallas, Texas

Solution:
Clearly, $n \geq 0$. If $n=0$, we get $n^2+2^n=1$, which is a perfect square. Let $n>0$. If $n$ is even, then $n=2k$ for some $k \in \mathbb{N}^*$. If $k \geq 7$, then $$(2^k)^2=2^{2k}<4k^2+2^{2k}<2^{2k}+2^{k+1}+1=(2^k+1)^2,$$ so $n^2+2^n$ is not a perfect square if $n$ is even and $n \geq 14$. So, $n \in \{2,4,6,8,10,12\}$. An easy check gives the solution $n=6$. If $n$ is odd, then $n=2k+1$ for some $k \in \mathbb{N}$. If $k=0$, we get no solutions, so assume $k \geq 1$. Let $m \in \mathbb{N}^*$ such that $(2k+1)^2+2^{2k+1}=m^2$. Then, $$(m-2k-1)(m+2k+1)=2^{2k+1}.$$ Since $m-2k-1<m+2k+1$ and the two factors have the same parity, then $$\begin{array}{lll} m-2k-1&=&2^a \\ m+2k+1&=&2^b, \end{array}$$ where $a,b \in \mathbb{N}$, $1 \leq a \leq b \leq 2k$ and $a+b=2k+1$. If $a \geq 2$, then subtracting we get $2(2k+1)=2^b-2^a=2^a(2^{b-a}-1)$, i.e. $2k+1=2^{a-1}(2^{b-a}-1)$, contradiction. So, $a=1$ and $b=2k$, which gives $2k+1=2^{2k-1}-1$, i.e. $k=2^{2k-2}-1$. If $k \geq 2$, then $k<2^{2k-2}-1$, so it must be $k=1$. But if $k=1$, we get no solutions. So, there are no solutions when $n$ is odd. We conclude that $n \in \{0,6\}$.

Mathematical Reflections 2017, Issue 1 - Problem J400

Problem:
Prove that for all real numbers $a,b,c$ the following inequality holds:
$$\dfrac{|a|}{1+|b|+|c|}+\dfrac{|b|}{1+|c|+|a|}+\dfrac{|c|}{1+|a|+|b|} \geq \dfrac{|a+b+c|}{1+|a+b+c|}.$$
When does the equality occur?

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
 Put $s=a+b+c$. By Triangle Inequality, we have $|s| \leq |a|+|b|+|c|$, so $|s|(1+|a|+|b|+|c|) \leq (1+|s|)(|a|+|b|+|c|)$, i.e.
$$\dfrac{|s|}{1+|s|} \leq \dfrac{|a|+|b|+|c|}{1+|a|+|b|+|c|}.$$ Using the fact that $|x| \geq 0$ for all real numbers $x$, we have
$$\begin{array}{lll} \dfrac{|a|+|b|+|c|}{1+|a|+|b|+|c|}&=&\dfrac{|a|}{1+|a|+|b|+|c|}+\dfrac{|b|}{1+|a|+|b|+|c|}+\dfrac{|c|}{1+|a|+|b|+|c|} \\ & \leq & \dfrac{|a|}{1+|b|+|c|}+\dfrac{|b|}{1+|c|+|a|}+\dfrac{|c|}{1+|a|+|b|}. \end{array}$$
So, $$\dfrac{|s|}{1+|s|} \leq \dfrac{|a|}{1+|b|+|c|}+\dfrac{|b|}{1+|c|+|a|}+\dfrac{|c|}{1+|a|+|b|},$$ which is the desired inequality. Equality occurs if and only if $|a|=|b|=|c|=0$, i.e. if and only if $a=b=c=0$.

Mathematical Reflections 2017, Issue 1 - Problem J399

Problem:
Two nine-digit numbers $m$ and $n$ are called cool if

   (a) they have the same digits but in different order,
   (b) no digit appears more than once,
   (c) $m$ divides $n$ or $n$ divides $m$.

Prove that if $m$ and $n$ are cool, then they contain digit $8$.

Proposed by Titu Andreescu, Dallas, Texas

Solution:
Assume by contradiction that there exist two \emph{cool} numbers $m$ and $n$ not containing digit $8$. Then in $m$ and $n$ appear the digits $0,1,2,3,4,5,6,7,9$ exactly once. Since the sum of their digits is $37$, then $m,n \equiv 1 \pmod{9}$. Assume without loss of generality that $m$ divides $n$. Then, $n=mk$, where $k$ is a natural number. Hence, $n-m=m(k-1)$. Reducing modulo $9$ this equation, we get $k-1 \equiv 0 \pmod{9}$, i.e. $k-1$ is divisible by $9$. Since $m$ and $n$ have the same digits in different order, then $m \neq n$, which gives $k \neq 1$. So, $k \geq 10$. But then $n \geq 10m$, i.e. $n$ has more digits than $m$, contradiction.

Mathematical Reflections 2017, Issue 1 - Problem J398

Problem:
Let $a,b,c$ be real numbers. Prove that $$(a^2+b^2+c^2-2)(a+b+c)^2+(1+ab+bc+ca)^2 \geq 0.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Let $x=a+b+c$ and $y=ab+bc+ca$. Then, $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=x^2-2y.$$ So,
$$\begin{array}{lll} (a^2+b^2+c^2-2)(a+b+c)^2+(1+ab+bc+ca)^2&=&(x^2-2y-2)x^2+(1+y)^2\\&=&x^4-2x^2(y+1)+(y+1)^2\\&=&(x^2-y-1)^2 \\ & \geq & 0. \end{array}$$
Equality holds if and only if $x^2=y+1$, i.e. if and only if $(a+b+c)^2=ab+bc+ca+1$, i.e. if and only if $a^2+b^2+c^2=1-ab-bc-ca$.

Mathematical Reflections 2017, Issue 1 - Problem J397

Problem:
Find all positive integers $n$ for which $3^4+3^5+3^6+3^7+3^n$ is a perfect square.

Proposed by Adrian Andreescu, Dallas, Texas

Solution:
We have $3^4+3^5+3^6+3^7+3^n=3240+3^n$. If $n \leq 4$, we obtain that $3240+3^n$ is a perfect square only when $n=2$. Assume that $n>4$. Then, there exists $t \in \mathbb{N}$ such that $3240+3^n=t^2$, i.e. $$3^4(40+3^{n-4})=t^2.$$ Since $3^4$ is a perfect square, it follows that $40+3^{n-4}$ must be a perfect square, so there exists $m \in \mathbb{N}$ such that $40+3^{n-4}=m^2$. If $n-4$ is odd, then $40+3^{n-4} \equiv 3 \pmod{4}$, so it cannot be a perfect square. It follows that $n-4$ is even, i.e. $n-4=2k$, where $k \in \mathbb{N}$, $k>2$. So, $40+3^{2k}=m^2$, i.e. $$(m-3^k)(m+3^k)=40.$$ Since $m-3^k<m+3^k$ and both factors have the same parity, then we obtain the two systems of equations
$$\begin{array}{lll} m-3^k&=&2 \\ m+3^k&=&20, \end{array} \qquad \begin{array}{lll} m-3^k&=&4 \\ m+3^k&=&10. \end{array}$$
Subtracting the first equation to the second equation of each system, we get $3^k=9$ or $3^k=3$, i.e. $k=2$ or $k=1$. So, $n=8$ or $n=6$. We conclude that $n \in \{2,6,8\}$.

Mathematical Reflections 2016, Issue 6 - Problem O391

Problem:
Find all 4-tuples $(x,y,z,w)$ of positive integers such that $$(xy)^3+(yz)^3+(zw)^3-252yz=2016.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
The given equation can be written as $$(xy)^3+(zw)^3+yz(y^2z^2-252)=2016.$$ Observe that it must be $yz(y^2z^2-252) \leq 2016$, so $yz \leq 18$. It follows that $$\begin{array}{rcl} (xy)^3+(zw)^3 & \in & \{2267,2512,2745,2960,3151,3312,3437,3520,3555,\\ & & 3536,3457,3312,3095,2800,2421,1952,1387,720\}. \end{array}$$
A case by case analysis gives $(xy)^3+(zw)^3 \in \{2745,2960\}$. If $(xy)^3+(zw)^3=2745$, we get $xy=1,zw=14,yz=3$ or $xy=14,zw=1,yz=3$, which gives no solutions. If $(xy)^3+(zw)^3=2960$, then $xy=6,zw=14,yz=4$ or $xy=14,zw=6,yz=4$. We get $(x,y,z,w) \in \{(3,2,2,7),(7,2,2,3)\}$.

Mathematical Reflections 2016, Issue 6 - Problem U395

Problem:
Evaluate $\displaystyle \int \dfrac{x^2+6}{(x \cos x-3\sin x)^2} \ dx$.

Proposed by Abdelouahed Hamdi, Doha, Qatar

Solution:Observe that $$\begin{array}{lll} \dfrac{x^2+6}{(x \cos x-3\sin x)^2}&=&\dfrac{x^2(\cos^2 x +\sin^2 x)+6(\cos^2 x + \sin^2 x)+5 x \sin x \cos x - 5x \sin x \cos x}{(x \cos x-3\sin x)^2} \\ &=& \dfrac{(x \cos x-2\sin x)(x \cos x-3 \sin x)+(x \sin x+3 \cos x)(x \sin x+2\cos x)}{(x\cos x-3\sin x)^2}\\&=& \dfrac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2} \\ &=&\left(\dfrac{f(x)}{g(x)}\right)', \end{array}$$ where $f(x)=x\sin x+3\cos x$ and $g(x)=x \cos x -3\sin x$. So, $$\int \dfrac{x^2+6}{(x \cos x-3\sin x)^2} \ dx=\dfrac{x\sin x+3\cos x}{x \cos x-3\sin x}+C.$$

Mathematical Reflections 2016, Issue 6 - Problem U391

Problem:
Find all positive integers $n$ such that $\varphi^3(n) \leq n^2$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:

Clearly $n=1$ satisfies the condition. Let $n>1$ and let $n=\prod_{j=1}^m p_j^{k_j}$, where $p_j$ is the $j$-th prime and $k_j \in \mathbb{N}$ for $j=1,2,\ldots,m$. Since $f(n)=n^2$ and $\varphi(n)$ are multiplicative functions, then the given relation becomes $$\prod_{j=1}^m \varphi^3(p_j^{k_j}) \leq \prod_{j=1}^m p_j^{2k_j}.$$ We use the following Lemma.


Lemma.

Let $p$ be a prime number and let $k$ be a positive integer.

    (a) If $k=0$, then $\varphi^3(p^k)=p^{2k}$.
    (b) If $p \geq 5$, then
    $$\varphi^3(p^k)>\dfrac{9}{4}p^{2k} \qquad \forall k \in \mathbb{N}^*. \qquad (1)$$
    (c) If $p \geq 7$, then $$\varphi^3(p^k)>4p^{2k} \qquad \forall k \in \mathbb{N}^*. \qquad (2)$$
    (d) If $p \geq 11$, then  $$\varphi^3(p^k)>8p^{2k} \qquad \forall k \in \mathbb{N}^*.  \qquad (3)$$

Proof. It's a simple computation.


From the inequality (1) in the Lemma we must consider only the primes $p_1=2$ and $p_2=3$.
If $k_1>3$, then $$\varphi^3(2^{k_1})=(2^{k_1}-2^{k_1-1})^3=2^{3(k_1-1)} > 2^{2k_1}$$ and if $k_2>1$, then $$\varphi^3(3^{k_2})=(3^{k_2}-3^{k_2-1})^3=3^{3(k_2-1)}\cdot8 > 3^{2k_2}.$$ So, there are no solutions if $k_1=k_2=0$ and $k_j>0$ for some $j>2$ or $k_1>3$ and $k_2>1$. So, $k_1 \leq 3$ or $k_2 \leq 1$.
\begin{description}

(a) If $k_1=0$, then $n=\prod_{j=2}^m p_j^{k_j}$. As before, we conclude immediately that there are solutions only if $k_2 \leq 1$ and $k_j=0$ for all $j=1,2,\ldots,m$. So, $n=3$.

(b) If $k_1=1$, then $n=2a$, where $a$ is an odd number. Hence, $$\varphi^3(n)=\varphi^3(2a)=\varphi^3(2)\varphi^3(a)=\varphi^3(a).$$ We have to find $a$ odd such that $$\varphi^3(a) \leq 4a^2.$$ From the inequality (2) in the Lemma we must consider only the primes $p_2=3$ and $p_3=5$. If $k_2>2$, then $$\varphi^3(3^{k_2})=3^{3(k_2-1)}\cdot2^3>4\cdot3^{2k_2}$$ and if $k_3>1$, then $$\varphi^3(5^{k_3})=5^{3(k_3-1)}\cdot4^3>4\cdot5^{2k_3}$$ So, there are no solutions for $a$ if $k_2=k_3=0$ and $k_j>0$ for some $j>3$ or $k_2>2$ and $k_3>1$. It follows that $k_2 \leq 2$ or $k_3 \leq 1$.

  (i) If $k_2=0$, then we have a solution if $k_3 \leq 1$, so $a \in \{1,5\}$, which gives $n \in \{2,10\}$.

  (ii) If $k_2=1$, then $a=3b$, where $b$ is a natural number nondivisible by $2$ and $3$. Hence, $n=6b$ and $$\varphi^3(n)=\varphi^3(6b)=\varphi^3(6)\varphi^3(b)=8\varphi^3(b).$$ We have to find $b$ nondivisible by $2$ and $3$ such that $8\varphi^3(b) \leq 36b^2$, i.e. $$\varphi^3(b) \leq \dfrac{9}{2}b^2.$$ From the inequality (3) in the Lemma we must consider only the primes $p_3=5$ and $p_4=7$. If $k_3>1$, then $$\varphi^3(5^{k_3})=5^{3(k_3-1)}\cdot4^3>\dfrac{9}{2}\cdot5^{2k_3}$$ and if $k_4>1$, then $$\varphi^3(7^{k_4})=7^{3(k_4-1)}\cdot6^3>\dfrac{9}{2}\cdot7^{2k_4}$$ So, there are no solutions for $b$ if $k_3=k_4=0$ and $k_j>0$ for some $j>4$ or $k_3>1$ and $k_4>1$. It follows that $k_3 \leq 1$ or $k_4 \leq 1$. If $k_3=0$, then we have a solution if $k_4 \leq 1$, so $b \in \{1,7\}$, which gives $n \in \{6,42\}$. If $k_3=1$, then $b=5c$, where $c$ is a natural number nondivisible by $2,3,5$. Hence, $n=30c$ and $$\varphi^3(n)=\varphi^3(30c)=\varphi^3(30)\varphi^3(c)=512\varphi^3(c).$$ We have to find $c$ nondivisible by $2,3,5$ such that $512\varphi^3(c) \leq 900c^2$, i.e. $$\varphi^3(c) \leq \dfrac{225}{128}c^2.$$ From the inequality (2) in the Lemma we have no primes for $c$. So, $c=1$ and $n=30$.

  (iii) If $k_2=2$, then $a=9b$, where $b$ is a natural number nondivisible by $2$ and $3$. Hence, $n=18b$ and $$\varphi^3(n)=\varphi^3(18b)=\varphi^3(18)\varphi^3(b)=216\varphi^3(b).$$ We have to find $b$ nondivisible by $2$ and $3$ such that $216\varphi^3(b) \leq 324b^2$, i.e. $$\varphi^3(b) \leq \dfrac{3}{2}b^2.$$ From the inequality (1) in the Lemma we have no primes for $b$. So, $b=1$ and $n=18$.

  (iv) If $k_3=0$, then we have a solution if $k_2 \leq 2$, so $a \in \{1,3,9\}$, which gives $n \in \{2,6,18\}$.

  (v) If $k_3=1$, then $a=5b$, where $b$ is a natural number nondivisible by $2$ and $5$. Hence, $n=10b$ and $$\varphi^3(n)=\varphi^3(10b)=\varphi^3(10)\varphi^3(b)=64\varphi^3(b).$$ We have to find $b$ nondivisible by $2$ and $5$ such that $64\varphi^3(b) \leq 100b^2$, i.e. $$\varphi^3(b) \leq \dfrac{25}{16}b^2.$$ From the inequality (2) in the Lemma we must consider only the prime $p_2=3$. If $k_2>1$, then $$\varphi^3(3^{k_2})=3^{3(k_2-1)}\cdot2^3>\dfrac{25}{16}\cdot3^{2k_2}.$$ So, there are no solutions for $b$ if $k_2=0$ and $k_j>0$ for some $j>3$ or $k_2>1$. It follows that $k_2 \leq 1$. If $k_2=0$, then $b=1$, which gives $n=10$. If $k_2=1$, then $b=3c$, where $c$ is a natural number nondivisible by $2,3,5$. Hence, $n=30c$ and we conclude as in point (ii).

(c) If $k_1=2$, then $n=4a$, where $a$ is an odd number. Hence, $$\varphi^3(n)=\varphi^3(4a)=\varphi^3(4)\varphi^3(a)=8\varphi^3(a).$$ We have to find $a$ odd such that $8\varphi^3(a) \leq 16a^2$, i.e. $$\varphi^3(a) \leq 2a^2.$$ From the inequality (1) in the Lemma we must consider only the prime $p_2=3$.  If $k_2>1$, then $$\varphi^3(3^{k_2})=3^{3(k_2-1)}\cdot2^3>2\cdot3^{2k_3}.$$ So, there are no solutions for $a$ if $k_2=0$ and $k_j>0$ for some $j>2$ or $k_2>1$. It follows that $k_2 \leq 1$. If $k_2=0$, then $a=1$ and $n=4$. If $k_2=1$, then $a=3b$, where $b$ is a natural number nondivisible by $2$ and $3$. Hence, $n=12b$ and $$\varphi^3(n)=\varphi^3(12b)=\varphi^3(12)\varphi^3(b)=64\varphi^3(b).$$ We have to find $b$ nondivisible by $2$ and $3$ such that $64\varphi^3(b) \leq 144b^2$, i.e. $$\varphi^3(b) \leq \dfrac{9}{4}b^2.$$ From the inequality (1) in the Lemma we have no primes for $b$. So, $b=1$ and $n=12$.

(d) If $k_1=3$, then $n=8a$, where $a$ is an odd number. Hence, $$\varphi^3(n)=\varphi^3(8a)=\varphi^3(8)\varphi^3(a)=64\varphi^3(a).$$ We have to find $a$ odd such that $64\varphi^3(a) \leq 64a^2$, i.e. $$\varphi^3(a) \leq a^2.$$ We know that this inequality has a solution if $k_2 \leq 1$, so $a \in \{1,3\}$, which gives $n \in \{8,24\}$.

(e) If $k_2=0$, then $n=\prod_{\substack{j=1 \\ j \neq 2}}^m p_j^{k_j}$. As before, we conclude immediately that there are solutions only if $k_1 \leq 3$ and $k_j=0$ for all $j=1,2,\ldots,m$. So, $n \in \{2,4,8\}$.

(f) If $k_2=1$, then $n=3a$, where $a$ is a natural number nondivisible by $3$. Hence, $$\varphi^3(n)=\varphi^3(3a)=\varphi^3(3)\varphi^3(a)=8\varphi^3(a).$$ We have to find $a$ nondivisible by $3$ such that $8\varphi^3(a) \leq 9a^2$, i.e. $$\varphi^3(a) \leq \dfrac{9}{8}a^2.$$ From the inequality (1) in the Lemma we must consider only the prime $p_1=2$. If $k_1>2$, then $$\varphi^3(2^{k_1})=2^{3(k_1-1)}>\dfrac{9}{8}\cdot2^{2k_1}.$$ So, there are no solutions for $a$ if $k_1$ and $k_j>0$ for some $j>2$ or $k_1>2$. It follows that $k_1 \leq 2$, so $a \in \{1,2,4\}$, which gives $n \in \{1,6,12\}$.

In conclusion, we have $n \in \{1,2,3,4,6,8,10,12,18,24,30,42\}$.

Note: Let $$A_x=\{n \in \mathbb{N}^* \ | \ \varphi(n) \leq n^x\}.$$ Observe that the function $$g(x)=|A_x|$$ is increasing for $x \in [0,1)$. The problem tells us that $g(2/3)=12$. Moreover, it's easy to see that $g(0)=2$ and $g(1/2)=4$.

Mathematical Reflections 2016, Issue 6 - Problem S395

Problem:
Let $a,b,c$ be positive integers such that $$a^2b^2+b^2c^2+c^2a^2-69abc=2016.$$
Find the least possible value of $\min(a,b,c)$.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Assume without loss of generality that $a \leq b \leq c$. If $a=1$, then $$b^2+b^2c^2+c^2-69bc=2016.$$ This equation is equivalent to
$$(2bc-67)^2+4(c-b)^2=12553.$$ It follows that $0 \leq c-b \leq 56$ and $0 < bc \leq 89$. Hence $b^2 \leq 89$, which gives $b \leq 9$. An easy check gives no positive integer solutions for $c$. If $a=2$, then $$4b^2+b^2c^2+4c^2-138bc=2016.$$
This equation is equivalent to $$(bc-65)^2+4(c-b)^2=6241.$$ It follows that $0 \leq c-b \leq 39$ and $0 < bc \leq 144$. Hence $b^2 \leq 144$, which gives $b \leq 12$. An easy check gives $b=c=12$, so the least possible value of $\min(a,b,c)$ is $2$.

Mathematical Reflections 2016, Issue 6 - Problem S393

Problem:
If $n$ is an integer such that $n^2+11$ is a prime, prove that $n+4$ is not a perfect cube.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Assume that $n+4$ is a perfect cube. Then $n+4=m^3$, where $m \in \mathbb{Z}$, i.e. $n=m^3-4$. It follows that
$$\begin{array}{lll} n^2+11&=&(m^3-4)^2+11\\&=&m^6-8m^3+27\\&=&m^6+m^3+27-9m^3\\&=&(m^2)^3+(m)^3+(3)^3-9m^3\\&=&(m^2+m+3)(m^4-m^3-2m^2-3m+9). \end{array}$$ As $m^2+m+3 \geq 3$ and $m^4-m^3-2m^2-3m+9 \geq 3$ for all $m \in \mathbb{Z}$, then  $n^2+11$ is not prime, contradiction.

Mathematical Reflections 2016, Issue 6 - Problem J331

Problem:
Solve the equation $$4x^3+\dfrac{127}{x}=2016.$$

Proposed by Adrian Andreescu, Dallas, Texas

Solution:
The given equation is equivalent to $$4x^4-2016x+127=0.$$ Observe that
$$\begin{array}{lll} 4x^4-2016x+127&=&4x^4-256x^2+(2x^2+16x)+127(2x^2-16x)+127\\&=&(2x^2-16x+1)(2x^2+16x)+127(2x^2-16x+1) \\ &=&(2x^2-16x+1)(2x^2+16x+127)\end{array}$$
So, the given equation becomes $$(2x^2-16x+1)(2x^2+16x+127)=0.$$
We obtain $$x=\dfrac{8 \pm \sqrt{62}}{2}, \qquad x=\dfrac{-8 \pm i\sqrt{190}}{2}.$$

Recreatii Matematice 2/2016, Problem VII.208

Problem:
Prove that the number $$N=2016^{n+1}-2015n-2016$$ has at least $27$ divisors for any $n \in \mathbb{N}^*$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
We have $$\begin{array}{lll} N&=&2016\cdot(2016^n-1)-2015n\\&=&2016\cdot2015\cdot(2016^{n-1}+2016^{n-2}+\ldots+1)-2015n\\&=&2015\cdot(2016^n+2016^{n-1}+\ldots+2016-n)\\&=&2015\cdot[(2016^n-1)+(2016^{n-1}-1)+\ldots+(2016-1)] \end{array}$$
Since $2016^n-1$ is divisible by $2015$ for all $n \geq 1$, we have that $2015^2 \ | \ N$. As $2015=5\cdot13\cdot31$, then $2015=5^2\cdot13^2\cdot31^2$, which has $27$ divisors. The conclusion follows.

Gazeta Matematica 6-7-8/2016, Problem 27253

Problem:
Evaluate $$\sum_{n=6}^{\infty} \dfrac{n^3-12n^2+47n-60}{n^5-5n^3+4n}$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that $$n^3-12n^2+47n-60=(n-5)(n-4)(n-3)$$ and $$n^5-5n^3+4n=(n-2)(n-1)n(n+1)(n+2),$$ so we have to evaluate
$$\sum_{n=6}^{\infty} \dfrac{(n-5)(n-4)(n-3)}{(n-2)(n-1)n(n+1)(n+2)}.$$
Set $f(x)=\dfrac{(x-5)(x-4)(x-3)}{(x-2)(x-1)x(x+1)(x+2)}$. Since $x=0,\pm 1,\pm 2$ are simple poles, then $$\textrm{Res}(f,2)=\lim_{x \to 2}(x-2)f(x)=-\dfrac{1}{4},$$ $$\textrm{Res}(f,1)=\lim_{x \to 1}(x-1)f(x)=4,$$ $$\textrm{Res}(f,0)=\lim_{x \to 0}xf(x)=-15,$$ $$\textrm{Res}(f,-1)=\lim_{x \to -1}(x+1)f(x)=20,$$ $$\textrm{Res}(f,-2)=\lim_{x \to -2}(x+2)f(x)=-\dfrac{35}{4}.$$
So, $$\begin{array}{lll} \dfrac{(n-5)(n-4)(n-3)}{(n-2)(n-1)n(n+1)(n+2)}&=&-\dfrac{1}{4(n-2)}+\dfrac{4}{n-1}-\dfrac{15}{n}+\dfrac{20}{n+1}-\dfrac{35}{4(n+2)}\\  &=&\left(-\dfrac{1}{4(n-2)}+\dfrac{1}{4(n+2)}\right)+\left(\dfrac{4}{n-1}-\dfrac{4}{n+1}\right)+\\&+&\left(-\dfrac{15}{n}+\dfrac{15}{n+1}\right)+\left(\dfrac{9}{n+1}-\dfrac{9}{n+2}\right). \end{array}$$
Since $$\sum_{n=6}^{\infty} \left(-\dfrac{1}{4(n-2)}+\dfrac{1}{4(n+2)}\right)=\dfrac{1}{4}\left(-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right),$$
$$\sum_{n=6}^{\infty} \left(\dfrac{4}{n-1}-\dfrac{4}{n+1}\right)=4\left(\dfrac{1}{5}+\dfrac{1}{6}\right),$$ $$\sum_{n=6}^{\infty} \left(-\dfrac{15}{n}+\dfrac{15}{n+1}\right)=-\dfrac{15}{6},$$ $$\sum_{n=6}^{\infty} \left(\dfrac{9}{n+1}-\dfrac{9}{n+2}\right)=-\dfrac{9}{8},$$ we have $$\sum_{n=6}^{\infty} \dfrac{(n-5)(n-4)(n-3)}{(n-2)(n-1)n(n+1)(n+2)}=\dfrac{1}{4}\left(-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right)+4\left(\dfrac{1}{5}+\dfrac{1}{6}\right)-\dfrac{15}{6}-\dfrac{9}{8}=\dfrac{1}{16}.$$

Gazeta Matematica 6-7-8/2016, Problem 27248

Problem:
Solve in real numbers the equation: $$\sqrt[3]{x^6-1}-\sqrt[3]{3x^5-5x^3+3x}=x^2-x-1.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that $$\begin{array}{lll} (x^6-1)-(3x^5-5x^3+3x)&=&x^6-3x^5+5x^3-3x-1\\&=&(x^2-x-1)(x^4-2x^3-x^2+2x+1)\\&=&(x^2-x-1)(x^2-x-1)^2\\&=&(x^2-x-1)^3. \end{array}$$ Then, from the identity $(a-b)^3=a^3-b^3-3ab(a-b)$, the given equation can be written as
$$(x^6-1)-(3x^5-5x^3+3x)-3\sqrt[3]{x^6-1}\cdot\sqrt[3]{3x^5-5x^3+3x}\cdot[\sqrt[3]{x^6-1}-\sqrt[3]{3x^5-5x^3+3x}]=(x^2-x-1)^3,$$ i.e. $$\sqrt[3]{x^6-1}\cdot\sqrt[3]{3x^5-5x^3+3x}\cdot[\sqrt[3]{x^6-1}-\sqrt[3]{3x^5-5x^3+3x}]=0.$$ Since $\sqrt[3]{x^6-1}-\sqrt[3]{3x^5-5x^3+3x}=x^2-x-1$, then the given equation becomes $$\sqrt[3]{x^6-1}\cdot\sqrt[3]{3x^5-5x^3+3x}\cdot(x^2-x-1)=0,$$ i.e. $$(x^6-1)(3x^5-5x^3+3x)(x^2-x-1)=0.$$ Finally, we have $$(x-1)(x^2+x+1)(x+1)(x^2-x+1)x(3x^4-5x^2+3)(x^2-x-1)=0.$$
Solving each equation, we get $x \in \left\{-1,0,1,\dfrac{1-\sqrt{5}}{2},\dfrac{1+\sqrt{5}}{2}\right\}$.

Gazeta Matematica 6-7-8/2016, Problem 27246

Problem:
Do there distinct prime numbers $p,q,r$ such that $\sqrt{p},\sqrt{q},\sqrt{r}$ are terms of an arithmetic progression?

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
The answer is no. Assume by contradiction that $\sqrt{p},\sqrt{q},\sqrt{r}$ are in the same arithmetic progression. If $d \neq 0$ is the common difference, then there exist integers $m,n \neq 0$ such that $$\begin{array}{rcl}\sqrt{q}-\sqrt{p}&=&md, \\ \sqrt{r}-\sqrt{p}&=&nd. \end{array}$$ Dividing the first and the second equation by $m$ and $n$ respectively and substituting, we obtain $$m(\sqrt{r}-\sqrt{p})=n(\sqrt{q}-\sqrt{p}),$$ whence $$m\sqrt{r}-n\sqrt{q}=(m-n)\sqrt{p}.$$ Squaring both sides, we obtain $$rm^2+qn^2-2mn\sqrt{rq}=p(m-n)^2,$$ which gives $$\sqrt{rq}=\dfrac{rm^2+qn^2-p(m-n)^2}{2mn}.$$ But this equality is false, since the left-hand side is irrational and the right-hand side is rational.

Crux Mathematicorum 2016, Issue 1 - Problem 4108

Problem:
Write $2010$ as a sum of consecutive squares. Is it possible to write $2014$ as the sum of several consecutive squares?

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
(a) Let $k$ be the number of consecutive perfect squares which satisfy the conditions and let $n \in \mathbb{N}$.  Then, $$(n+1)^2+(n+2)^2+\ldots+(n+k)^2=2010.$$ Since $$\begin{array}{lll}(n+1)^2+\ldots+(n+k)^2&=&kn^2+2n(1+2+\ldots+k)+(1^2+2^2+\ldots+k^2)\\&=&kn^2+2n\cdot\dfrac{k(k+1)}{2}+\dfrac{k(k+1)(2k+1)}{6}\\&=&\dfrac{k}{6}[6n^2+6n(k+1)+(k+1)(2k+1)], \end{array}$$ we get
$$k[6n^2+6n(k+1)+(k+1)(2k+1)]=12060.    \qquad          (1)$$
Moreover, since $\dfrac{k(k+1)(2k+1)}{6}=1^2+2^2+\ldots+k^2 \leq (n+1)^2+(n+2)^2+\ldots+(n+k)^2=2010$, we get $k \leq 17$. So, $k \ | \ 12060$ and $k \leq 17$, which gives $k \in \{1,2,3,4,5,6,9,10,12,15\}$. Observe that if $k=4h+2$, where $h \in \mathbb{N}$, then the left-hand side in (1) is not divisible by $4$, but the right-hand side is divisible by $4$. So, $k \in \{1,3,5,9,12,15\}$. If $k=15$, equation (1) gives $$6n(n+16)+496=804 \implies 6n(n+16)=308,$$ a contradiction. If $k=12$, equation (1) gives $$6n(n+13)+325=1005 \implies 6n(n+13)=680,$$ a contradiction. If $k=9$, equation (1) gives $$6n(n+10)+190=1340 \implies 6n(n+10)=1150,$$ a contradiction. If $k=5$, equation (1) gives $$6n(n+6)+66=2412 \implies n(n+6)=391,$$ which gives $n=17$. Therefore, $18^2+19^2+20^2+21^2+22^2=2010$, and the maximum number of consecutive perfect squares which satisfy the condition is $5$.

(b) The answer is no. Indeed, if there exist $k,n \in \mathbb{N}$ such that $$(n+1)^2+(n+2)^2+\ldots+(n+k)^2=2014,$$ then as we have seen in (a),
$$k[6n^2+6n(k+1)+(k+1)(2k+1)]=2014\cdot6=12084   \qquad        (2)$$
and $\dfrac{k(k+1)(2k+1)}{6} \leq 2014$. So, $k \ | \ 12084$, $k \leq 17$ and $k \neq 4h+2$ for any $h \in \mathbb{N}$. It follows that $k \in \{1,3,4,12\}$. Since $2014$ is not a perfect square, $k \neq 1$. As $24^2+25^2+26^2=1877<2014<25^2+26^2+27^2=2030$, then $k \neq 3$. As $20^2+21^2+22^2+23^2=1854<2014<21^2+22^2+23^2+24^2=2030$, then $k \neq 4$. Finally, if $k=12$ equation (2) gives $6n(n+13)+325=1007$, i.e. $6n(n+13)=682$, a contradiction. Therefore there does not exist $k \in \mathbb{N}$ which satisfy the condition, and the conclusion follows.