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 $(x,y,z,u,v,w)=(2,3,5,7,13,37)$.