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 $x=0$.