Gazeta Matematica 4/2014, Problem E:14645

Problem:
Show that for any integer $n > 3$, the number $a_n=10(n^2+1)$ can be written as the sum of four distinct perfect squares.

Proposed by Alessandro Ventullo

Solution:
We have $$\begin{eqnarray*} 10(n^2+1)&=&8n^2+2+2n^2+8\\&=&4n^2-4n+1+4n^2+4n+1+n^2-4n+4+n^2+4n+4\\&=&(2n-1)^2+(2n+1)^2+(n-2)^2+(n+2)^2, \end{eqnarray*}$$ and the statement follows. 

Gazeta Matematica 4/2014, Problem E:14641

Problem:
Find all primes $p$ such that the equation $$x^3+y^3+3xy=p^2+1$$ has integer solutions.

Proposed by Alessandro Ventullo

Solution:
Observe that $x^3+y^3-1+3xy=(x+y-1)(x^2+y^2+1-xy+x+y)$. So, the given equation can be rewritten as
$$(x+y-1)(x^2+y^2+1-xy+x+y)=p^2.$$ Since $x^2+y^2+1-xy+x+y=\dfrac{1}{2}[(x-y)^2+(x+1)^2+(y+1)^2] \geq 0$, then $x+y-1\geq0$, i.e. $x+y \geq 1$. This gives $x^2+y^2+1-xy+x+y\geq 2|xy|+1-xy+1=2|xy|-xy+2 \geq 2$. Moreover, $x^2+y^2+1-xy+x+y>x+y-1$. So, it must be
$$\begin{array}{rll} x+y-1&=&1 \\ x^2+y^2+1-xy+x+y&=&p^2. \end{array}$$
The first equation gives

$x+y=2$                                (1)

the second equation gives $(x+y)^2-3xy+(x+y)+1=p^2$, i.e.

$xy=\dfrac{7-p^2}{3}.$                                (2)

So, $$t^2=\left(\dfrac{x-y}{2}\right)^2=\dfrac{(x+y)^2-4xy}{4}=1-\dfrac{7-p^2}{3},$$ i.e. $3t^2=p^2-4$. If $p>2$, we have that $3t^2 \equiv 0,3 \pmod{4}$ and $p^2-4 \equiv 1 \pmod{4}$, a contradiction. So, $p=2$. This value is acceptable because, from equations (1) and (2), we get $(x,y)=(1,1)$.

Mathematical Reflections 2014, Issue 5 - Problem O313

Problem:
Find all positive integers $n$ for which there are positive integers $a_0, a_1, \ldots, a_n$ such that $a_0 + a_1 + \ldots + a_n = 5(n-1)$ and $$\dfrac{1}{a_0}+\dfrac{1}{a_1}+\ldots+\dfrac{1}{a_n}=2.$$

Proposed by Titu Andreescu

Solution:
Assume without loss of generality that $a_0 \leq a_1 \leq \ldots \leq a_n$. By the AM-GM Inequality, we have $$10(n-1)=\left(a_0 + a_1 + \ldots + a_n\right)\left(\dfrac{1}{a_0}+\dfrac{1}{a_1}+\ldots+\dfrac{1}{a_n}\right) \geq (n+1)^2,$$ so $n \leq 6$. Clearly, $n>1$. We have five cases.

(i) $n=2$. We immediately get $a_0=1,a_1=2,a_2=2$.

(ii) $n=3$. We immediately get $a_0=1,a_1=3,a_2=3,a_3=3$.

(iii) $n=4$. Observe that $\dfrac{5}{a_0}\geq 2$, so $a_0=1,2$. If $a_0=1$, then $a_1+a_2+a_3+a_4=14$ and $\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}=1$, but this would imply $$14=(a_1+a_2+a_3+a_4)\left(\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}\right)\geq 16,$$ contradiction. So, $a_0=2$, which gives $a_1+a_2+a_3+a_4=13$ and $\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}=\dfrac{3}{2}$. Now, $\dfrac{4}{a_1} \geq \dfrac{3}{2}$ implies that $a_1=2$. So, $a_2+a_3+a_4=11$ and $\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}=1$. Since $\dfrac{3}{a_2} \geq 1$, we get $a_2=2$ or $a_2=3$. If $a_2=2$, then $a_3+a_4=9$ and $\dfrac{1}{a_3}+\dfrac{1}{a_4}=\dfrac{1}{2}$, which gives $a_3=3,a_4=6$. If $a_2=3$, then $a_3+a_4=8$ and $\dfrac{1}{a_3}+\dfrac{1}{a_4}=\dfrac{2}{3}$, so no solutions.

(iv) $n=5$. Reasoning as before, we get $a_0=2,a_1=2,a_2=4,a_3=4,a_4=4,a_5=4$.

(v) $n=6$. Observe that $\dfrac{7}{a_0} \geq 2$, so $a_0 \leq 3$. But if $a_0=i$ for $i=1,2$, then $$(25-i)\left(2-\dfrac{1}{i}\right)=\left(a_1+\ldots+a_6\right)\left(\dfrac{1}{a_1}+\ldots+\dfrac{1}{a_6} \right) \geq 36$$ gives a contradiction. So, $a_0=3$, which gives $a_1+\ldots+a_6=22$ and $\dfrac{1}{a_1}+\ldots+\dfrac{1}{a_6}=\dfrac{5}{3}$. Therefore, $a_1=3$, which gives $a_2+\ldots+a_6=19$ and $\dfrac{1}{a_2}+\ldots+\dfrac{1}{a_6}=\dfrac{4}{3}$. Then $a_2=3$, which gives $a_3+a_4+a_5+a_6=16$ and $\dfrac{1}{a_3}+\dfrac{1}{a_4}+\dfrac{1}{a_5}+\dfrac{1}{a_6}=1$, so $a_3=4,a_4=4,a_5=4,a_6=4$.

We conclude that $n=2,3,4,5,6$.

Mathematical Reflections 2014, Issue 5 - Problem U315

Problem:
Let $X$ and $Y$ be complex matrices of the same order with $XY^2-Y^2X=Y$. Prove that $Y$ is nilpotent.

Proposed by Radouan Boukharfane.

Solution:
We prove by induction on $k \in \mathbb{Z}^+$ that $\textrm{tr}(Y^k)=0$ for all $k \in \mathbb{Z}^+$, so that the conclusion will follow. We have
$$\textrm{tr}(Y)=\textrm{tr}(XY^2-Y^2X)=\textrm{tr}(XY^2)-\textrm{tr}(Y^2X)=\textrm{tr}(Y^2X)-\textrm{tr}(Y^2X)=0.$$
Assume that $\textrm{tr}(Y^k)=0$ for some $k \in \mathbb{Z}^+$. Then,
$$\begin{array}{lll}\textrm{tr}(Y^{k+1})&=&\textrm{tr}(Y^k\cdot Y)\\&=&\textrm{tr}(Y^k(XY^2-Y^2X))\\&=&\textrm{tr}((Y^kX)Y^2))-\textrm{tr}(Y^{k+2}X)\\&=&\textrm{tr}(Y^{k+2}X)-\textrm{tr}(Y^{k+2}X)\\&=&0, \end{array}$$ as desired.

Mathematical Reflections 2014, Issue 5 - Problem U314

Problem:
Prove that for any positive integer $k$,
$$\lim_{n \to \infty} \left(\dfrac{1+\sqrt[n]{2}+\ldots+\sqrt[n]{k}}{k}\right)>\dfrac{k}{e},$$
where $e$ is Euler constant.

Proposed by Ivan Borsenco.

 

Solution:
Let $f(x)=\sqrt[n]{x}$ and let $\mathcal{P}=\{[0,1],\ldots,[k-1,k]\}$ be a partition of the interval $I=[0,k]$. Since $f$ is strictly increasing on $I$, we have $$\dfrac{1}{k}\sum_{i=1}^k \sqrt[n]{i}> \dfrac{1}{k}\int_{0}^k \sqrt[n]{x} \ dx=\dfrac{n}{n+1}\sqrt[n]{k}.$$ Therefore,
$$\left(\dfrac{1}{k}\sum_{i=1}^k \sqrt[n]{i}\right)^n>k\left(1-\dfrac{1}{n+1}\right)^n$$ and taking the limit of both sides, we obtain
$$\lim_{n \to \infty}\left(\dfrac{1}{k}\sum_{i=1}^k \sqrt[n]{i}\right)^n>\dfrac{k}{e}.$$

Mathematical Reflections 2014, Issue 5 - Problem U313

Problem:
Let $X$ and Y be nonnegative definite Hermitian matrices such that $X-Y$ is also non-negative definite. Prove that $\textrm{tr}(X^2) \geq \textrm{tr}(Y^2)$.

Proposed by Radouan Boukharfane

Solution:
We use Klein's Inequality:
For all nonnegative-definite Hermitian $n \times n$ matrices $A$ and $B$, and all differentiable convex functions $f:\mathbb{R}^+ \longrightarrow \mathbb{R}$, the following inequality holds:
$${\rm tr}[f(A)- f(B)- (A - B)f'(B)] \geq 0.$$

Take $A=X, B=Y$ and $f(x)=x^2$. By Klein's Inequality, we get $$\textrm{tr}\left(X^2-Y^2-(X-Y)\cdot2Y\right) \geq 0.$$ Moreover, since $X$ and $X-Y$ are nonnegative, then $$\textrm{tr}\left((X-Y)Y\right) \geq 0.$$ This can be proved by Cholesky decomposition: $X-Y=L_1L_1^*$ and $Y=L_2L_2^*$ for some lower triangular matrices $L_1,L_2$ with real and nonnegative diagonal entries. So, $$\textrm{tr}\left((X-Y)Y\right)=\textrm{tr}(L_1L_1^*L_2L_2^*)=\textrm{tr}(L_2^*L_1L_1^*L_2)=\textrm{tr}((L_1^*L_2)^*(L_1^*L_2))\geq 0.$$ Hence,
$$\begin{array}{lll}\textrm{tr}\left(X^2\right)-\textrm{tr}\left(Y^2\right)&=&\textrm{tr}\left(X^2-Y^2\right)\\&=&\textrm{tr}\left(X^2-Y^2-(X-Y)\cdot2Y\right)+2\textrm{tr}\left((X-Y)Y\right) \geq 0, \end{array}$$ which is the desired conclusion.

Mathematical Reflections 2014, Issue 5 - Problem J318

Problem:
Determine the functions $f : \mathbb{R} \longrightarrow \mathbb{R}$ satisfying $f(x-y)-xf(y) \leq 1-x$ for all real numbers $x$ and $y$.

Proposed by Marcel Chirita

Solution:
Observe that if $x=0$, we get $f(-y) \leq 1$ for all $y \in \mathbb{R}$, i.e. $f(x) \leq 1$ for all $x \in \mathbb{R}$. If $y=0$, then

$f(x)-xf(0) \leq 1-x,$                               (1)

and if $y=x$, then

$f(0)-xf(x) \leq 1-x.$                               (2)

Summing up (1) and (2), we get $(1-x)(f(x)+f(0)) \leq 1-x$ and if $x>1$, we obtain $f(x)+f(0) \geq 2$, for all $x \in (1,\infty)$, i.e. $f(x) \geq 2-f(0) \geq 1$. Therefore, $f(x)=1$ for all $x \in (1,\infty)$ and $f(0)=1$. Now, setting $x=1$ in the given relation, we have $f(1-y) \leq f(y)$ and for all $1-y>1$, we have $1 \leq f(y)$. So, $f(x)=1$ for all $x \in (-\infty,0)$. Now, setting $x=1$ in (1) and (2), we get $f(0)=f(1)$. Finally, if we set $x=1/t$ and $y=x$ in the given relation, we get $$f(0)-\dfrac{1}{t}f\left(\dfrac{1}{t}\right) \leq 1-\dfrac{1}{t},$$ i.e. $$tf(0)-f\left(\dfrac{1}{t}\right) \leq t-1.$$ This relation, $f(0)=1$ and (1) give $$f(x) \leq f\left(\dfrac{1}{x}\right),$$ so $f(x)=1$ for all $x \in (0,1)$. In conclusion, $f(x)=1$ for all $x \in \mathbb{R}$. 

Mathematical Reflections 2014, Issue 5 - Problem J316

Problem:
Solve in prime numbers the equation $$x^3+y^3+z^3+u^3+v^3+w^3=53353.$$

Proposed by Titu Andreescu.

Solution:
Assume $x \leq y \leq z \leq u \leq v \leq w$. Since $53353$ is odd, then $x=2$. We have
$$y^3+z^3+u^3+v^3+w^3=53345.$$ Moreover, $5w^3 \geq 53345 \geq w^3$, so $w \in \{23,29,31,37\}$. A case by case analysis gives $(x,y,z,u,v,w)=(2,3,5,7,13,37)$.

Mathematical Reflections 2014, Issue 4 - Problem U312

Problem:
Let $p$ be a prime and let $R$ be a commutative ring with characteristic $p$. Prove that the
sets $S_k = \{x \in R \ | \ x^p=k\}$, where $k \in \{1,\ldots,p\}$, have the same number of elements.

Proposed by Corneliu Manescu-Avram.

Solution:
If for any $x \in R$ and $k \in \{1,\ldots,p\}$ it holds $x^p \neq k$, then there is nothing to prove. Assume that there is some $x \in R$ such that $x^p=k$ for some $k \in \{1,\ldots,p\}$. Hence, $k \in R$. Consider the set $S=\{x+k \ | \ x \in R\}$. Observe that the mapping $\varphi:R \longrightarrow S$ defined by $\varphi(x)=x+k$ is bijective. Indeed, if $\varphi(x)=\varphi(y)$, then $x+k=y+k$, i.e. $x=y$ and this proves that $\varphi$ is injective. If $y \in S$, then $y=x+k$ for some $x \in R$, so it's enough to take $x=y-k \in R$ in order to have $y=\varphi(x)$, and this proves that $\varphi$ is surjective. Let $n \in \{1,\ldots,p\}$ and consider the set $S_n = \{x \in R \ | \ x^p=n\}$. Since $(x+k)^p=x^p+k^p=x^p+k$, then $$\varphi(S_n)=\begin{cases} S_{n-k} & \textrm{if } n>k \\ S_{p-k+n} & \textrm{if } k \leq n. \end{cases}$$ It follows that $|S_n|=|S_{n-k}|$ or $|S_n|=|S_{p-k+n}|$. By arbitrarily of $n \in \{1,\ldots,p\}$, we conclude that $|S_1|=|S_2|=\ldots=|S_p|$.

Mathematical Reflections 2014, Issue 4 - Problem U307

Problem:
Prove that any polynomial $f \in \mathbb{R}[X]$ can be written as a difference of increasing polynomials.

Proposed by Jishnu Bose.

Solution:
Let $f(X)=a_nX^n+a_{n-1}X^{n-1}+\ldots+a_1X+a_0$. We construct the following algorithm. Set $g_0(X)=a_0, h_0(X)=0$ and for $k=1,\ldots,n$, if $k$ is odd, then $$g_k(X)=\begin{cases} a_kX^k & \textrm{if } a_k \geq 0 \\ 0 & \textrm{if } a_k<0, \end{cases}$$ $$h_k(X)=\begin{cases} 0 & \textrm{if } a_k \geq 0 \\ -a_kX^k & \textrm{if } a_k<0, \end{cases}$$ and if $k$ is even, then $$g_k(X)=\begin{cases}a_k(X^{k+1}+X^k+X^{k-1}) & \textrm{if } a_k \geq 0 \\ -a_k(X^{k+1}+X^{k-1}) & \textrm{if } a_k < 0,  \end{cases}$$ $$h_k(X)=\begin{cases}a_k(X^{k+1}+X^{k-1}) & \textrm{if } a_k \geq 0 \\ -a_k(X^{k+1}+X^k+X^{k-1}) & \textrm{if } a_k < 0. \end{cases}$$
Observe that for all $k=0,\ldots,n$, we have $g_k(X)-h_k(X)=a_kX^k$ and $g'_k(X) \geq 0$, $h'_k(X) \geq 0$ for all $x \in \mathbb{R}$. Setting $$g(X)=\sum_{k=0}^n g_k(X), \qquad h(X)=\sum_{k=0}^n h_k(X),$$ we have that $g(X)$ and $h(X)$ are increasing polynomials and $f(X)=g(X)-h(X)$.

Mathematical Reflections 2014, Issue 4 - Problem S307

Problem:
Let $ABC$ be a triangle such that $\angle ABC - \angle ACB = 60^{\circ}$. Suppose that the length of the
altitude from $A$ is $\dfrac{1}{4}BC$. Find $\angle ABC$.

 Proposed by Omer Cerrahoglu and Mircea Lascu.

Solution:
Let $\alpha=\angle ABC$ and let $H$ be the foot of the altitude from $A$. Observe that $\alpha>60^{\circ}$, so $\tan \alpha>\sqrt{3}$. We have $$\dfrac{AH}{BH}=\tan \alpha, \qquad \dfrac{AH}{CH}=\tan (\alpha-60^{\circ}).$$ Since $AH=\dfrac{1}{4}BC$, then $$BC=BH+CH=\dfrac{1}{4 \tan \alpha}BC+\dfrac{1}{4 \tan (\alpha-60^{\circ})}BC,$$ which gives $$\dfrac{1}{\tan \alpha}+\dfrac{1}{\tan (\alpha-60^{\circ})}=4.$$ Since $\dfrac{1}{\tan (\alpha)-60^{\circ}}=\dfrac{1+\tan 60^{\circ}\tan \alpha}{\tan \alpha-\tan 60^{\circ}}$, setting $x=\tan \alpha$, we get $$\dfrac{1}{x}+\dfrac{1+\sqrt{3}x}{x-\sqrt{3}}=4 \iff (4-\sqrt{3})x^2-2(2\sqrt{3}+1)x+\sqrt{3}=0.$$ As $x>\sqrt{3}$, then $x=2+\sqrt{3}$, i.e. $\alpha=75^{\circ}$. 

Mathematical Reflections 2014, Issue 4 - Problem J311

Problem:
Let $a, b, c$ be real numbers greater than or equal to $1$. Prove that
$$\dfrac{a(b^2+3)}{3c^2+1}+\dfrac{b(c^2+3)}{3a^2+1}+\dfrac{c(a^2+3)}{3b^2+1} \geq 3.$$

Proposed by Titu Andreescu.

Solution:
First, observe that $x \geq 1$ is equivalent to $$(x-1)^3 \geq 0 \iff x^3+3x \geq 3x^2+1 \iff \dfrac{x(x^2+3)}{3x^2+1} \geq 1.$$
By the AM-GM Inequality, we have
$$\dfrac{a(b^2+3)}{3c^2+1}+\dfrac{b(c^2+3)}{3a^2+1}+\dfrac{c(a^2+3)}{3b^2+1} \geq 3\sqrt[3]{\dfrac{a(a^2+3)}{3a^2+1}\cdot\dfrac{b(b^2+3)}{3b^2+1}\cdot\dfrac{c(c^2+3)}{3c^2+1}} \geq 3,$$ which is the desired conclusion.

Mathematical Reflections 2014, Issue 4 - Problem J309

Problem:

Let $n$ be an integer greater than $3$ and let $S$ be a set of $n$ points in the plane that are not the vertices of a convex polygon and such that no three are collinear. Prove that there is a triangle with the vertices among these points having exactly one other point from $S$ in its interior.

Proposed by Ivan Borsenco.

Solution:

We use the following theorem.

Carathéodory's Theorem.
If $\mathbf{x} \in \mathbb{R}^d$ lies in the convex hull of a set $S$, there is a subset $S'$ of $S$ consisting of $d+1$ or fewer points such that $\mathbf{x}$ lies in the convex hull of $S'$.

Let $\mathbf{x} \in S$ such that $\mathbf{x}$ lies in the interior of the convex hull of $S$. Hence, by Caratheodory's Theorem ($d=2$), there is a subset $S'$ of $S$ having at most $3$ points such that $\mathbf{x}$ lies in the convex hull of $S'$. Since the points are not collinear, then $|S'| \neq 2$ and since $\mathbf{x}$ lies in the interior of the convex hull of $S$, then $|S'| \neq 1$. So, $|S'|=3$ and there exists a triangle $T$ whose vertices are in $S$ which contains $\mathbf{x}$. If $\mathbf{x}$ is the unique point of $S$ in $T$, we are done. If there is another point $\mathbf{y} \neq \mathbf{x}$ of $S$ in $T$, joining $\mathbf{x}$ with the three points in $S'$ we have that $\mathbf{y}$ lies in the interior of one among these triangles. Proceeding in this way and using the fact that $S$ has finitely many points, then there exists a triangle whose vertices are in $S$ which has exactly one point of $S$ in its interior.

Mathematical Reflections 2014, Issue 4 - Problem J308

Problem:
Are there triples $(p, q, r)$ of primes for which $(p^2-7)(q^2-7)(r^2-7)$ is a perfect square?

Proposed by Titu Andreescu.

Solution:
The answer is no. Let $N=(p^2-7)(q^2-7)(r^2-7)$. If exactly one or exactly three among $p,q,r$ are equal to $2$, then $N<0$ and if exactly two numbers are equal to $2$, then $N \equiv 2 \pmod{4}$. In every case $N$ is not a perfect square. Assume that $p,q,r$ are all odd. Then there exist $a,b,c \in \mathbb{N}$ such that $p^2=4a+1,q^2=4b+1,r^2=4c+1$ and
$$N=2^3(2a-3)(2b-3)(2c-3).$$ Since in the factorization of $N$ the number $2$ has an odd exponent, we conclude that $N$ is not a perfect square. Therefore, $N$ is never a perfect square.

Mathematical Reflections 2014, Issue 4 - Problem J307

Problem:
Prove that for each positive integer $n$ there is a perfect square whose sum of digits is equal to $4^n$.

Proposed by Mihaly Bencze.

Solution:
Consider the sequence $$\begin{array}{rcl}a_1&=&4 \\ \qquad a_n&=&\underbrace{11\ldots1}_{n-1 \textrm{ times}}0\underbrace{22\ldots2}_{n-1 \textrm{ times}}4, \end{array} \qquad n \geq 2.$$
Every term of the sequence is a perfect square. Indeed, $a_0$ is clearly a perfect square and $$\begin{array}{lll} a_n&=&\dfrac{10^{n-1}-1}{9}\cdot10^{n+1}+2\cdot\dfrac{10^{n-1}-1}{9}\cdot10+4\\ &=&\dfrac{10^{2n}-8\cdot10^n+16}{9}\\&=&\left(\dfrac{10^n-4}{3}\right)^2, \end{array}$$ where $n \geq 2$.
Furthermore, if $s$ denotes the sum of the digits of $a_n$, we have $s(a_1)=4$ and $s(a_n)=(n-1)+2(n-1)+4=3n+1$ for all $n \geq 2$. Since $4^n \equiv 1 \pmod{3}$ for all $n \in \mathbb{N}$, we have that for all $n \in \mathbb{N}$ there exists $k \in \mathbb{N}$ such that $4^n=3k+1=s(a_k)$ and the conclusion follows.

Mathematical Reflections 2014, Issue 2 - Problem J299

Problem:
Prove that no matter how we choose $n$ numbers from the set $\{1, 2, \ldots, 2n\}$, one of them
will be a square-free integer.



Solution:
Let $A=\{1,2,\ldots,2n\}$. We will prove that there are at most $n-1$ elements in $A$ which are not square-free integers. The number of the elements in $A$ which are divisible by $p^2$, where $p$ is a prime number, are
$$\left\lfloor \dfrac{2n}{p^2} \right\rfloor \leq \dfrac{2n}{p^2}.$$ Therefore, the number of elements in $A$ which are not square-free integers are $$\sum_{p=2 \atop p \textrm{ prime}}^{n} \left\lfloor \dfrac{2n}{p^2} \right\rfloor \leq \sum_{p=2 \atop p \textrm{ prime}}^{n} \dfrac{2n}{p^2} < \sum_{i=2}^{n} \dfrac{2n}{i^2} < \sum_{i=2}^{n} \dfrac{2n}{i(i-1)}=2n\left(\dfrac{1}{2}-\dfrac{1}{n(n-1)}\right)<n,$$ and the conclusion follows.

Mathematical Reflections 2014, Issue 2 - Problem U297

Problem:
Let $a_0 = 0, a_1 = 2$, and $a_{n+1}=\sqrt{2-\dfrac{a_{n-1}}{a_n}}$ for $n \geq 0$. Find $\displaystyle \lim_{n \to \infty} 2^n a_n$.

Proposed by Titu Andreescu.

Solution:
We will show by induction on $n$ that $a_n=2 \sin \dfrac{\pi}{2^n}$ for all $n \in \mathbb{N}$. If $n=0$, it's obvious. Suppose that $a_n=2 \sin \dfrac{\pi}{2^n}$ for some $n \in \mathbb{N}$. Then, $$a_{n+1}=\sqrt{2-\dfrac{2\sin (\pi/2^{n-1})}{2\sin (\pi/2^n)}}=\sqrt{2-\dfrac{4\sin (\pi/2^n)\cos (\pi/2^n)}{2\sin (\pi/2^n)}}=\sqrt{2-2\cos \dfrac{\pi}{2^n}}=2\sin \dfrac{\pi}{2^{n+1}}.$$
Therefore, $$2^n a_n=2^{n+1}\sin\dfrac{\pi}{2^n}=2^{n+1}\left(\dfrac{\pi}{2^n}+o((\pi/2^n)^2)\right),$$ which implies $$\lim_{n \to \infty} 2^n a_n=2\pi.$$

Mathematical Reflections 2014, Issue 2 - Problem U295

Problem:
Let $a$ be a real number such that $(\lfloor na \rfloor)_{n \geq 1}$ is an arithmetic sequence. Prove that $a$ is an integer.

Proposed by Mihai Piticari.

Solution:
Let $a_n=\lfloor na \rfloor$ and $b_n=n|a|$. By hypothesis, there exists $d \in \mathbb{R}$ such that $a_{n+1}-a_n=d$. Observe that $d \in \mathbb{Z}$ since it is the difference of two integers. Now, $(b_n)_{n \geq 1}$ is positive, increasing and unbounded and $$\lim_{n \to \infty} \dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}=\dfrac{d}{|a|}.$$ By the Stolz-Cesaro Theorem, we obtain $\displaystyle \lim_{n \to \infty} \dfrac{a_n}{b_n}=\dfrac{d}{|a|}$, which gives $$\lim_{n \to \infty} \dfrac{\lfloor na \rfloor}{n}=d.$$ Moreover, since $na-1 < \lfloor na \rfloor \leq na$, we have $$\lim_{n \to \infty} \dfrac{na-1}{n} \leq \lim_{n \to \infty} \dfrac{\lfloor na \rfloor}{n} \leq \lim_{n \to \infty} \dfrac{na}{n},$$ and by the Squeeze Theorem, we obtain $\displaystyle \lim_{n \to \infty} \dfrac{\lfloor na \rfloor}{n}=a$. By uniqueness of the limit, we obtain $a=d$ and the conclusion follows.

Mathematical Reflections 2014, Issue 2 - Problem J296

Problem:
Several positive integers are written on a board. At each step, we can pick any two
numbers $u$ and $v$, where $u \geq v$, and replace them with $u + v$ and $u - v$. Prove that after
a finite number of steps we can never obtain the initial set of numbers.

Proposed by Marius Cavachi.

Solution:
Let $S$ be the sum of the numbers on the blackboard at a given moment. After removing the numbers $u$ and $v$ and replacing them with $u+v$ and $u-v$ we have that the sum of the numbers on the blackboard is $$S'=S-u-v+(u+v)+(u-v)=S+(u-v)>S.$$ Therefore, after each step the sum of the numbers on the blackboard increases, and this proves that after a finite number of step we can never obtain the original numbers.

Mathematical Reflections 2014, Issue 2 - Problem J295

Problem:
Let $a,b,c$ be positive integers such that $(a-b)^2+(b-c)^2+(c-a)^2=6abc$. Prove that $a^3+b^3+c^3+1$ is not divisible by $a+b+c+1$.

Proposed by Mihaly Bencze.

Solution:
Clearly $$(a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2-ab-bc-ac)$$ which implies $$a^2+b^2+c^2-ab-bc-ac=3abc.$$
Now, using the identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$ we obtain that $$a^3+b^3+c^3-3abc=3abc(a+b+c),$$ or $$a^3+b^3+c^3=3abc(a+b+c+1)$$ which means that $a+b+c+1$ divides $a^3+b^3+c^3$. If $a+b+c+1$ divides $a^3+b^3+c^3+1$, then $a+b+c+1$ divides $(a^3+b^3+c^3+1)-(a^3+b^3+c^3)=1$, which is impossible, since $a+b+c+1>1$. Thus, $a^3+b^3+c^3+1$ is not divisible by $a+b+c+1$.

Mathematical Reflections 2014, Issue 1 - Problem O289

Problem:
Let $a, b, x, y$ be positive real numbers such that $x^2-x+1=a^2$, $y^2+y+1=b^2$, and
$(2x-1)(2y+1)=2ab+3$. Prove that $x+y=ab$.

Proposed by Titu Andreescu.

Solution:
Multiplying both sides of the first two equations by $4$ and both sides of the third equation by $2$ , we have
$$\begin{array}{rll} (2x-1)^2+3&=&4a^2            (1)\\ (2y+1)^2+3&=&4b^2            (2)\\ 2(2x-1)(2y+1)&=&4ab+6     (3)\\ \end{array}$$
Summing up these equations, we get $$(2x-1)^2+2(2x-1)(2y+1)+(2y+1)^2+6=4(a^2+ab+b^2)+6,$$ i.e. $$[(2x-1)+(2y+1)]^2=4(a^2+ab+b^2),$$ which gives $x+y=\sqrt{a^2+ab+b^2}$. So, it suffices to show that $\sqrt{a^2+ab+b^2}=ab$. Multiplying $(1)$ and $(2)$, we get $[(2x-1)^2+3][(2y+1)^2+3]=16a^2b^2$, and using $(3)$ we get $$[(2ab+3)^2+3(4a^2-3+4b^2-3)+9]=16a^2b^2,$$ which gives $a^2+ab+b^2=a^2b^2$. Taking the square root, we obtain the conclusion.

Mathematical Reflections 2014, Issue 1 - Problem U294

Problem:
Let $p_1, p_2, \ldots, p_n$ be pairwise distinct prime numbers. Prove that
$$\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})=\mathbb{Q}(\sqrt{p_1}+\sqrt{p_2}+\ldots+\sqrt{p_n}).$$

Proposed by Marius Cavachi.

Solution:
Clearly, $\mathbb{Q}(\sqrt{p_1}+\sqrt{p_2}+\ldots+\sqrt{p_n})$ is a subfield of $\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})$. Observe that $\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})$ is Galois over $\mathbb{Q}$, since it is the splitting field of the polynomial $(x^2-\sqrt{p_1})\cdots(x^2-\sqrt{p_n})$. Every automorphism $\sigma$ is completely determined by its action on $\sqrt{p_1},\ldots,\sqrt{p_n}$, which must be mapped to $\pm \sqrt{p_1},\ldots,\pm \sqrt{p_n}$, respectively. Therefore, $\textrm{Gal}(\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})/\mathbb{Q})$ is the group generated by $\{\sigma_1,\sigma_2,\ldots,\sigma_n\}$, where $\sigma_i$ is the automorphism defined by $$\sigma_i(\sqrt{p_j})=\begin{cases} -\sqrt{p_j} & \textrm{if } i=j \\ \sqrt{p_j} & \textrm{if } i \neq j. \end{cases}$$ Clearly, the only automorphism that fixes $\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})$ is the identity. Moreover, it's easy to see that the only automorphism that fixes the element $\sqrt{p_1}+\ldots+\sqrt{p_n}$ is the identity, which means that the only automorphism that fixes $\mathbb{Q}(\sqrt{p_1}+\ldots+\sqrt{p_n})$ is the identity. Hence, by the Fundamental Theorem of Galois Theory, $\mathbb{Q}(\sqrt{p_1}+\sqrt{p_2}+\ldots+\sqrt{p_n})=\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_n})$.

Mathematical Reflections 2014, Issue 1 - Problem U289

Problem:
Let $a \geq 1$ be such that $\left(\left\lfloor a^n \right\rfloor \right)^{\frac{1}{n}} \in \mathbb{Z}$ for all sufficiently large integers $n$. Prove that $a \in \mathbb{Z}$.

Proposed by Mihai Piticari.

Solution:
Suppose that $\left(\lfloor a^n \rfloor \right)^{\frac{1}{n}}=m_n$ for all $n \geq n_0$, where $n_0 \in \mathbb{Z}^+$ and $m_n \in \mathbb{Z}$. Hence, $\lfloor a^n \rfloor=m_n^n$, so for all $n \geq n_0$ we have $a^n-1 < m_n^n \leq a^n$, i.e. $$\sqrt[n]{a^n-1} < m_n \leq a \qquad \forall n \geq n_0.$$
By the Squeeze Theorem, since $\displaystyle \lim_{n \to \infty} \sqrt[n]{a^n-1}=\lim_{n \to \infty} a=a$, we have that $\displaystyle \lim_{n \to \infty} m_n=a$. Since $\{m_n\}_{n \geq n_0}$ is a convergent sequence of integers, therefore must be constant for all $n \geq n_1 \geq n_0$. Indeed, $\displaystyle \lim_{n \to \infty} m_n=a$ implies that there exists $n_1 \in \mathbb{Z}^+$ such that for all $n \geq n_1$ we have $a-1/2<m_n<a+1/2$. Since $(a-1/2,a+1/2)$ has length $1$, it follows that there is only $m_n$ in this interval. This means that $m_n$ is constant for all $n \geq n_1$ and so $m_n=a$ for all $n \geq n_1$, which gives $a \in \mathbb{Z}$.

Mathematical Reflections 2014, Issue 1 - Problem S294

Problem:
Let $s(n)$ be the sum of digits of $n^2+1$. Define the sequence $(a_n)_{n \geq 0}$ by $a_{n+1} = s(a_n)$, with $a_0$ an arbitrary positive integer. Prove that there is $n_0$ such that $a_{n+3} = a_n$ for all $n \geq n_0$.

Proposed by Roberto Bosch Cabrera.

Solution:
We have to prove that the given sequence is $3$-periodic. Since $f(5)=8, f(8)=11$ and $f(11)=5$, it suffices to prove that for every positive integer $a_0$ there exists some $n \in \mathbb{N}$ such that $a_n \in \{5,8,11\}$. Let $m$ be the number of digits of $a_0$. We prove the statement by induction on $m$. For $m \leq 2$ we proceed by a direct check. If $a_0 \in \{5,8,11\}$ there is nothing to prove. If $a_0$ is a two-digit number, then $a^2_0 \leq 10000$, so $a_1 \leq 37$ and we reduce to analyze the cases for $a_0 \leq 37$.

(i) If $a_0 \in \{2,7,20\}$, then $a_1=5$. If $a_0 \in \{1,10,26,28\}$, then $a_1 \in \{2,20\}$, so $a_2=5$. Finally, if $a_0 \in \{3,6,9,12,15,18,27,30,33\}$, then $a_3=5$.
(ii) If $a_0 \in \{4,13,23,32\}$, then $a_1=8$.
(iii) If $a_0 \in \{17,19,21,35,37\}$, then $a_1=11$. If $a_0 \in \{14,22,24,31,36\}$, then $a_1 \in \{17,19\}$, so $a_2=11$. Finally, if $a_0 \in \{16,25,29,34\}$, then $a_3=11$.

Thus, we have proved that if $a_0$ is a one or two-digit number, then $a_n \in \{5,8,11\}$ for some $n \in \mathbb{N}$, i.e. the sequence is $3$-periodic. Let $m \geq 2$ and suppose that the statement is true for all $k \leq m$. Let $a_0$ be an $(m+1)$-digit number. Then, $10^m \leq a_0 < 10^{m+1}$ which implies $10^{2m} \leq a^2_0 < 10^{2(m+1)}$. Hence, $$a_1=f(a_0) \leq 9\cdot2(m+1)+1<10^m,$$ and by induction hypothesis, the sequence $(a_n)_{n \geq 1}$ is $3$-periodic, which implies that the sequence $(a_n)_{n \geq 0}$ is $3$-periodic, as we wanted to prove. 

Mathematical Reflections 2014, Issue 1 - Problem S290

Problem:
Prove that there is no integer $n$ for which $$\dfrac{1}{2^2}+\dfrac{1}{3^2}+\ldots+\dfrac{1}{n^2}=\left(\dfrac{4}{5}\right)^2.$$

Proposed by Ivan Borsenco.

Solution:
Let $p$ be the greatest prime number such that $p \leq n < 2p$. Then the given equality can be written as $$5^2k=(4\cdot n!)^2,$$ where $k=\displaystyle \sum_{i=1}^n \dfrac{(n!)^2}{i^2}$. Observe that $k \equiv \dfrac{(n!)^2}{p^2} \not \equiv 0 \pmod{p}$. Since $p|5^2k$ and $p$ does not divide $k$, it follows that $p|5^2$, i.e. $p=5$. So, $n \in \{5,6,7,8,9\}$. An easy check shows that none of these values satisfies the equality.

Mathematical Reflections 2014, Issue 1 - Problem J293

Problem:
Find all positive integers $x, y, z$ such that $$(x+y^2+z^2)^2-8xyz=1.$$

Proposed by Aaron Doman.

Solution:
We rewrite the equation as $$x^2+2x(y^2+z^2-4yz)+(y^2+z^2)^2-1=0.$$ Since $x$ must be a positive integer, the discriminant of this quadratic equation in $x$ must be non-negative, i.e. $$(y^2+z^2-4yz)^2-(y^2+z^2)^2+1 \geq 0,$$ which is equivalent to $$-8yz(y-z)^2+1 \geq 0,$$ which gives $yz(y-z)^2 \leq 1/8$. Since $y$ and $z$ are positive integers, it follows that $$yz(y-z)^2=0,$$ so $y-z=0$, i.e. $y=z$. The given equation becomes $(x-2y^2)^2=1$, which yields $x=2y^2 \pm 1$. Therefore, all the positive integer solutions to the given equation are $$(2n^2-1,n,n), (2n^2+1,n,n), \qquad n \in \mathbb{Z}^+.$$

Mathematical Reflections 2014, Issue 1 - Problem J292

Problem:
Find the least real number $k$ such that for every positive real numbers $x, y, z$, the following
inequality holds: $$\prod_{cyc} (2xy+yz+zx) \leq k(x+y+z)^6.$$

Proposed by Dorin Andrica.

Solution:
Observe that $$(x-y)^2+z^2 \geq 0 \iff x^2+y^2+z^2-2xy \geq 0 \iff (x+y+z)^2 \geq 4xy+2yz+2zx,$$ so $2xy+yz+zx \leq \dfrac{1}{2}(x+y+z)^2$. Likewise, $2yz+zx+xy \leq \dfrac{1}{2}(x+y+z)^2$ and $2zx+xy+yz \leq \dfrac{1}{2}(x+y+z)^2$. Therefore,
$$\prod_{cyc} (2xy+yz+zx) \leq \dfrac{1}{8}(x+y+z)^6.$$ The equality is attained if and only if $(x-y)^2+z^2=0, (y-z)^2+x^2=0, (z-x)^2+y^2=0$, i.e. if and only if $x=y=z=0$. Clearly, $k_{\min}=1/8$, because for each inequality $1/2$ was the least real number for which the inequality was satisfied.


Note. Actually, this is not the minimum value of $k$, which is $64/729$. See the official solution.

Mathematical Reflections 2014, Issue 1 - Problem J290

Problem:
Let $a, b, c$ be nonnegative real numbers such that $a + b + c = 1$. Prove that
$$\sqrt[3]{13a^3+14b^3}+\sqrt[3]{13b^3+14c^3}+\sqrt[3]{13c^3+14a^3} \geq 3.$$

Proposed by Titu Andreescu.

Solution:

The second Minkowski's Inequality (see \emph{Zdravko Cvetkovski - Inequalities. Theorems, Techniques and Selected Problems, page 99}) states that if $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$ are positive real numbers and $p>1$, then
$$\left(\left(\sum_{i=1}^n a_i\right)^p+\left(\sum_{i=1}^n b_i\right)^p \right)^{1/p} \leq \sum_{i=1}^n (a^p_i+b^p_i)^{1/p}.$$
Equality occurs if and only if $\dfrac{a_1}{b_1}=\dfrac{a_2}{b_2}=\ldots=\dfrac{a_n}{b_n}$.\\
Now, take $$(a_1,a_2,a_3)=(\sqrt[3]{13}a,\sqrt[3]{13}b,\sqrt[3]{13}c),$$ $$(b_1,b_2,b_3)=(\sqrt[3]{14}b, \sqrt[3]{14}c,\sqrt[3]{14}a)$$ and $p=3$. Then,
$$\begin{array}{lll} \sqrt[3]{13a^3+14b^3}+\sqrt[3]{13b^3+14c^3}+\sqrt[3]{13c^3+14a^3} & \geq & \sqrt[3]{\left(\sqrt[3]{13(a+b+c)}\right)^3+\left(\sqrt[3]{14(a+b+c)}\right)^3} \\ & = & \sqrt[3]{(\sqrt[3]{13})^3+(\sqrt[3]{14})^3} \\ &=& 3. \end{array}$$
Equality occurs if and only if $\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}$. 

Mathematical Reflections 2014, Issue 1 - Problem J289

Problem:
Let $a$ be a real number such that $0 \leq a < 1$. Prove that
$$\left\lfloor a\left(1+\left\lfloor \dfrac{1}{1-a} \right\rfloor \right)\right\rfloor+1=\left\lfloor \dfrac{1}{1-a} \right\rfloor.$$

Proposed by Arkady Alt.

Solution:

Let $m \in \mathbb{N}$ and $\alpha \in [0,1)$ such that $\dfrac{1}{1-a}=m+\alpha$. Then, we have to prove that
$$\left\lfloor \left(1-\dfrac{1}{m+\alpha}\right)(1+m) \right\rfloor+1=m.$$
Indeed, $$\left\lfloor \left(1-\dfrac{1}{m+\alpha}\right)(1+m) \right\rfloor=\left\lfloor m+1-\dfrac{m+1}{m+\alpha} \right\rfloor=m-1,$$ where the last equality follows from $-2<-\dfrac{m+1}{m+\alpha}<-1$.