Mathematical Reflections 2016, Issue 5 - Problem O387

Problem:
Are there integers $n$ for which $$3^{6n-3}+3^{3n-1}+1$$ is a perfect cube?

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

Solution:
Clearly if $n \leq 0$, the number $3^{6n-3}+3^{3n-1}+1$ is never a perfect cube. Let $n>0$. Then,
$$(3^{2n-1})^3=3^{6n-3} < 3^{6n-3}+3^{3n-1}+1 < 3^{6n-3}+3^{4n-1}+3^{2n}+1=(3^{2n-1}+1)^3.$$
So, there are no integers $n$ such that $3^{6n-3}+3^{3n-1}+1$ is a perfect cube.

Mathematical Reflections 2016, Issue 5 - Problem O386

Problem:
Find all pairs $(m,n)$ of positive integers such that $3^m-2^n$ is a perfect square.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Let $n \in \mathbb{N}$ such that
$$3^x-2^y=n^2 \qquad (1).$$
We have three cases.

(i) If $y \geq 2$, then $3^x \equiv n^2 \pmod{4}$, so $x$ must be even, i.e. $x=2k$ for some $k \in \mathbb{N}^*$. Therefore, equation (1) becomes $$(3^k-n)(3^k+n)=2^y.$$ It follows that $3^k-n=2^a$ and $3^k+n=2^b$, where $a,b \in \mathbb{N}$ and $a+b=y$. Moreover, $a<b$ and adding these two equations, we get $$2\cdot3^k=2^a+2^b.$$ If $a=0$, the LHS is even and the RHS is odd, contradiction. If $a \geq 2$, we obtain $2\cdot3^k \equiv 0 \pmod{4}$, contradiction. So, $a=1$ and $b=y-1$ and we get $$3^k=1+2^{y-2}.$$ If $y=3$, we obtain $k=1$, i.e. $x=2$. If $y \geq 4$, then $3^k \equiv 1 \pmod{4}$, so $k$ is even and we can write $$(3^{\frac{k}{2}}-1)(3^{\frac{k}{2}}+1)=2^{y-2}.$$ Since $3^{\frac{k}{2}}+1$ and $3^{\frac{k}{2}}-1$ are powers of $2$ and their difference is $2$, we obtain $3^{\frac{k}{2}}+1=4$ and $3^{\frac{k}{2}}-1=2$, i.e. $k=2$, which gives $x=4$ and $y=5$. Therefore, we obtain the solutions $(x,y) \in \{(2,3),(4,5)\}$.

(ii) If $y=1$, then $3^x-2=n^2$. We have that $x$ must be odd, otherwise $n^2 \equiv -1 \pmod{4}$, contradiction. In the ring of integers $\mathbb{Z}[\sqrt{-2}]$, we have $$3^x=(n-\sqrt{-2})(n+\sqrt{-2}).$$
Let $d=(n-\sqrt{-2},n+\sqrt{-2})$. Clearly, $d \ | \ 2\sqrt{-2}$, so $N(d) \ | \ 8$. Since $3^x$ is odd, then its norm is odd and this implies that $N(d)=1$, i.e. $d=\pm 1$. So, $n-\sqrt{-2}$ and $n+\sqrt{-2}$ are coprime in $\mathbb{Z}[\sqrt{-2}]$. Since these two factors have the same norm and the only non trivial factorization (up to sign permutations) of $3$ with factors with the same norm is $3=(1-\sqrt{-2})(1+\sqrt{-2})$, then this forces
$$\begin{array}{rcl} n-\sqrt{-2}&=&(1-\sqrt{-2})^x \\ n+\sqrt{-2}&=&(1+\sqrt{-2})^x  \end{array} \qquad \textrm{or} \qquad \begin{array}{rcl} n-\sqrt{-2}&=&(1+\sqrt{-2})^x \\ n+\sqrt{-2}&=&(1-\sqrt{-2})^x  \end{array}$$
Considering the first equation, we get $\textrm{Im}(n-\sqrt{-2})=\textrm{Im}((1-\sqrt{-2})^x)$, i.e. $$-\sqrt{2}=\left[-{x \choose 1}+2{x \choose 3}-4{x \choose 5}+\ldots+(-1)^{\frac{x+1}{2}}2^{\frac{x-1}{2}}{x \choose x}\right]\sqrt{2},$$ i.e. $$-1=-{x \choose 1}+2{x \choose 3}-4{x \choose 5}+\ldots+(-1)^{\frac{x+1}{2}}2^{\frac{x-1}{2}}{x \choose x}.$$ Let $$\displaystyle f(x)=\sum_{k \textrm{ odd}}^x {x \choose k} (-1)^{\frac{k+1}{2}}2^{\frac{k-1}{2}}$$ be defined on the odd natural numbers. An easy check shows that $f(x) \leq f(x+2)$ for any odd $x$. Since $f(5)=11$, then $x \in \{1,3\}$. An easy check shows that $f(1)=f(3)=-1$ and we get $(x,y) \in \{(1,1),(3,1)\}$. If we consider the second system of equations, we get $\textrm{Im}(n-\sqrt{-2})=\textrm{Im}((1+\sqrt{-2})^x)$ and proceeding as before, we obtain no solutions.

(iii) If $y=0$, then $3^x-1=n^2$, i.e. $3^x=n^2+1$. Since $x>0$ and $n^2 \equiv 0,1 \pmod{3}$, we get no solutions in this case.

In conclusion, $(x,y) \in \{(1,1),(2,3),(3,1),(4,5)\}$.

Mathematical Reflections 2016, Issue 5 - Problem O385

Problem:
Let $f(x,y)=\frac{x^3-y^3}{6}+3xy+48$. Let $m$ and $n$ be odd integers such that $$|f(m,n)| \leq mn+37.$$ Evaluate $f(m,n)$

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

Solution:
Let $m$ and $b$ be two odd integers such that $$-mn-37 \leq \dfrac{m^3-n^3}{6}+3mn+48 \leq mn+37.$$ We get the two inequalities
$$m^3-n^3+64+12mn \leq -2$$ and $$m^3-n^3+512+24mn \geq 2.$$ Using the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ and setting $(a,b,c)=(m,-n,4)$ and $(a,b,c)=(m,-n,8)$, we obtain the two inequalities
$$(m-n+4)(m^2+n^2+16+mn-4m+4n) \leq -2$$
and
$$(m-n+8)(m^2+n^2+64+mn-8m+8n) \geq 2$$
Since $a^2+b^2+c^2-ab-bc-ca \geq 0$ with the equality if and only if $a=b=c$, then and as $m,n$ are odd integers, we obtain that $m^2+n^2+16+mn-4m+4n>0$ and $m^2+n^2+16+mn-4m+4n>0$. It follows that $m-n+4<0$ and $m-n+8>0$, i.e. $m-n+4 \leq -2$ and $m-n+8 \geq 2$, which gives $-6 \leq m-n \leq -6$, i.e. $m-n=-6$. Therefore, $$f(m,n)=-(m^2+mn+n^2)+3mn+48=-(m-n)^2+48=12.$$

Mathematical Reflections 2016, Issue 5 - Problem S389

Problem:
Let $n$ be a positive integer. Prove that for any integers $a_1, a_2, \ldots, a_{2n+1}$ there is a rearrangement $b_1, b_2, \ldots, b_{2n+1}$ such that $2^n\cdot n!$ divides $$(b_1-b_2)(b_3-b_4)\cdot \ldots \cdot (b_{2n-1}-b_{2n}).$$

Proposed by Cristinel Mortici, Valahia University, Targoviste, Romania

Solution:
Observe that we have $2n$ remainders modulo $2n$. By the pigeonhole principle there exist two integers among $a_1,a_2,\ldots,a_{2n+1}$, say $b_1$ and $b_2$, such that $(b_1-b_2)$ is divisible by $2n$. Now, discard $b_1$ and $b_2$. There remain $2n-1$ numbers among the ones in the given list. Since we have $2n-2$ remainders modulo $2n-2$, by the pigeonhole principle we have that there exist two integers, say $b_3$ and $b_4$, such that $(b_3-b_4)$ is divisible by $2n-2$. Proceeding in this way, we obtain a rearrangement $b_1, b_2, \ldots, b_{2n+1}$ of $a_1, a_2, \ldots, a_{2n+1}$ such that $$(b_1-b_2)(b_3-b_4)\cdot \ldots \cdot(b_{2n-1}-b_{2n})$$ is divisible by $2n\cdot (2n-2) \cdot (2n-4) \cdot \ldots \cdot 4 \cdot 2=2^n \cdot n!$.

Mathematical Reflections 2016, Issue 5 - Problem S385

Problem:
Let $a,b,c$ be positive real numbers. Prove that $$\dfrac{1}{a^3+8abc}+\dfrac{1}{b^3+8abc}+\dfrac{1}{c^3+8abc} \leq \dfrac{1}{3abc}.$$

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

Solution:
The given inequality can be written as $$\dfrac{1}{9abc}-\dfrac{1}{a^3+8abc}+\dfrac{1}{9abc}-\dfrac{1}{b^3+8abc}+\dfrac{1}{9abc}-\dfrac{1}{c^3+8abc} \geq 0,$$ i.e. $$\dfrac{a^2-bc}{a^2+8bc}+\dfrac{b^2-ca}{b^2+8ca}+\dfrac{c^2-ab}{c^2+8ab} \geq 0.$$
Let $x=\dfrac{bc}{a^2}$, $y=\dfrac{ca}{b^2}$, $z=\dfrac{ab}{c^2}$. Then, we have $xyz=1$ and we want to prove that $$\dfrac{1-x}{1+8x}+\dfrac{1-y}{1+8y}+\dfrac{1-z}{1+8z} \geq 0.$$ Observe that $$\dfrac{1-x}{1+8x}+\dfrac{1-y}{1+8y}+\dfrac{1-z}{1+8z}=\dfrac{3(-64xyz+16(xy+yz+zx)+5(x+y+z)+1)}{(1+8x)(1+8y)(1+8z)}.$$ Since $xyz=1$, then by AM-GM Inequality we have $x+y+z \geq 3$ and $xy+yz+zx \geq 3$, so $$-64xyz+16(xy+yz+zx)+5(x+y+z)+1 \geq -63+16\cdot3+5\cdot3=0$$ and we get
$$\dfrac{1-x}{1+8x}+\dfrac{1-y}{1+8y}+\dfrac{1-z}{1+8z} \geq 0,$$ as we wanted to prove.

Mathematical Reflections 2016, Issue 5 - Problem J389

Problem:
Solve in real numbers the system of equations
$$(x^2-y+1)(y^2-x+1)=2[(x^2-y)^2+(y^2-x)^2]=4.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Set $u=x^2-y$ and $v=y^2-x$. The given system can be rewritten as $$\begin{array}{rcl} u^2+v^2&=&2 \\ (u+1)(v+1)&=&4. \end{array}$$
Multiplying the second equation by $2$ and adding to the first equation, we get $$(u+v)^2+2(u+v)-8=0,$$ and solving for $u+v$, we get $u+v=-4$ or $u+v=2$. If $u+v=-4$, then $uv=4-u-v-1=7$. If $u+v=2$, then $uv=4-u-v-1=1$. We obtain the two systems of equations:
$$\begin{array}{rcl} u+v&=&-4 \\ uv&=&7, \end{array} \qquad \begin{array}{rcl} u+v&=&2 \\ uv&=&1. \end{array}$$
The first system gives no real solution. The second system gives $u=v=1$. We obtain the system of equations
$$\begin{array}{rcl} x^2-y&=&1 \\ y^2-x&=&1.\end{array}$$ Subtracting side by side the two equations, we get $(x^2-y^2)+(x-y)=0$, i.e. $$(x-y)(x+y+1)=0.$$ So, $x=y$ or $x=-1-y$. Substituting these values, we obtain $$(x,y) \in \left\{\left(\dfrac{1+\sqrt{5}}{2},\dfrac{1+\sqrt{5}}{2}\right),\left(\dfrac{1-\sqrt{5}}{2},\dfrac{1-\sqrt{5}}{2}\right),(0,-1),(-1,0),\right\}.$$

Mathematical Reflections 2016, Issue 5 - Problem J387

Problem:
Find all digits $a,b,c,x,y,z$ for which $\overline{abc},\overline{xyz}$ and $\overline{abcxyz}$ are all perfect squares (no leading zeros allowed).

Proposed by Adrian Andreescu, Dallas, Texas

Solution:
Let $\overline{abcxyz}=n^2$, where $n \in \mathbb{N}$. Since $\overline{abc}$ must be a perfect square, it's easy to see that $n \in \{317,348,349,380,412,443,475,506,538,570,601,633,696,728,759,791,854,886,949\}$. An easy check shows that the given conditions are satisfied if and only if $n \in \{380,475,570\}$. So, $$(a,b,c,x,y,z) \in \{(1,4,4,0,0,0),(2,2,5,6,2,5),(3,2,4,9,0,0)\}.$$

Mathematical Reflections 2016, Issue 5 - Problem J386

Problem:
Find all real solutions to the system of equations $$x+yzt=y+ztx=z+txy=t+xyz=2.$$

Proposed by Mohamad Kouroshi, Tehran, Iran

Solution:
Subtracting the second equation to the first equation, the third equation to the second equation, the fourth equation to the third equation and the first equation to the fourth equation, we obtain
$$\begin{array}{rcl} (x-y)(1-zt)&=&0 \\ (y-z)(1-tx)&=&0 \\ (z-t)(1-xy)&=&0 \\ (t-x)(1-yz)&=&0. \end{array}$$

We have four cases.

(i) $x-y=0$ and $z-t=0$, i.e. $x=y$ and $z=t$. Substituting these values into the second and the fourth equation, we get $(x-z)(1-zx)=0$. If $x=z$, then $x+x^3=2$, which gives $x=y=z=t=1$. If $zx=1$, then $y+t=2$, i.e. $x+z=2$, which gives $x=1,z=1$, so $x=y=z=t=1$.

(ii) $x-y=0$ and $1-xy=0$, i.e. $x=y$ and $xy=1$. We obtain $x=y=\pm 1$. If $x=y=1$, then $zt=1$ and $z+t=2$, which gives $z=t=1$. If $x=y=-1$, then $zt=-3$ and $z+t=2$, which gives $z=3,t=-1$ or $z=-1,t=3$.

(iii) $1-zt=0$ and $z-t=0$, i.e. $z=t$ and $zt=1$. We obtain $z=t=\pm 1$. If $z=t=1$, then $xy=1$ and $x+y=2$, which gives $x=y=1$. If $z=t=-1$, then $xy=-3$ and $x+y=2$, which gives $x=3,y=-1$ or $x=-1,y=3$.

(iv) $1-zt=0$ and $1-xy=0$, i.e. $zt=1$ and $xy=1$. We obtain $x+y=2$ and $z+t=2$, which gives $x=y=z=t=1$.

In conclusion, the real solutions to the given system of equations are $$(x,y,z,t) \in \{(1,1,1,1),(1,1,3,-1),(1,1,-1,3),(3,-1,1,1),(-1,3,1,1)\}.$$

Mathematical Reflections 2016, Issue 5 - Problem J385

Problem:
 If the equalities $$\begin{array}{rcl} 2(a+b)-6c-3(d+e)&=&6 \\ 3(a+b)-2c+6(d+e)&=&2 \\ 6(a+b)+3c-2(d+e)&=&-3 \end{array}$$ hold simultaneously, evaluate $a^2-b^2+c^2-d^2+e^2$.

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

Solution:
Multiplying the second equation by $3$ and subtracting the first equation, we get $$a+b=-3(d+e).$$ Multiplying the third equation by $2$ and adding the first equation, we get $$2(a+b)=d+e.$$ It follows that $a+b=-6(a+b)$, which gives $a+b=0$ and $d+e=0$. So, $c=-1$ and
$$a^2-b^2+c^2-d^2+e^2=(a-b)(a+b)+c^2-(d-e)(d+e)=1.$$

Mathematical Reflections 2016, Issue 4 - Problem U381

Problem:
Find all positive integers $n$ such that
$$\sigma(n)+d(n)=n+100.$$

(We denoted by $\sigma(n)$ the sum of the divisors of $n$ and by $d(n)$ the number of divisors of $n$).

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Since $\sigma(n) \geq n+1$, then the sum of the proper divisors of $n$ is $99-d(n) \leq 97$. Denoting by $\omega(n)$ the number of distinct primes dividing $n$, observe that if $\omega(n) \geq 4$, then there exist four prime numbers $p<q<r<s$ that divide $n$. Since $p \geq 2$, $q \geq 3$, $r \geq 5$ and $s \geq 7$, then $qrs$ divides $n$ but $qrs \geq 3\cdot5\cdot7>97$, which contradicts the hypothesis. So, $\omega(n) \leq 3$. We have three cases.

(i) $\omega(n)=1$. Then, $n=p^k$, where $p$ is a prime number and $k \in \mathbb{N}^*$. If $k=1$, then $(1+p)+2=p+100$, contradiction. If $k \geq 2$, then $$1+p+\ldots+p^k+(k+1)=p^k+100,$$ i.e.
$$p(1+p+\ldots+p^{k-2})+k=98. \qquad (1)$$
If $k \geq 7$, then $$p(1+p+\ldots+p^{k-2}) \geq 2(1+2+\dots+2^5)=2\cdot63>97.$$ It follows that $k \in \{2,3,4,5,6\}$. An easy check in (1) shows that there are no solutions.

(ii) $\omega(n)=2$. Then, $n=p^{k_1}q^{k_2}$, where $p,q$ are prime numbers, $p<q$ and $k_1,k_2 \in \mathbb{N}^*$. Since $p \geq 2$ and $q \geq 3$, if $n=p^5q$, then $p^5+p^4q+p^3q \geq 32+16\cdot3+8\cdot3>97$, so $k_1 \leq 4$. If $n=pq^4$, then $q^4+q^3 \geq 81+27>97$, so $k_2 \leq 3$.  We have twelve cases.

(a) $k_1=1,k_2=1$. Then, $n=pq$ and $d(n)=4$. It follows that $p+q=95$. By parity, it must be $p=2$, but $q=93$ is not a prime. So, no solutions in this case.

(b) $k_1=2,k_2=1$. Then, $n=p^2q$ and $d(n)=6$. It follows that $p+p^2+q+pq=93$. By parity, it must be $p=2$ and we get $q=29$. So, $n=116$.

(c) $k_1=3, k_2=1$. Then, $n=p^3q$ and $d(n)=8$. It follows that $p+p^2+p^3+q+pq+p^2q=91$. By parity, it must be $p=2$ and we get $q=11$. So, $n=88$.

(d) $k_1=4, k_2=1$. Then, $n=p^4q$ and $d(n)=10$. It follows that $p+p^2+p^3+p^4+q+pq+p^2q+p^3q=89$. By parity, it must be $p=2$ and we get $q=59/11$, i.e. no solution.

(e) $k_1=1, k_2=2$. Then, $n=pq^2$ and $d(n)=6$. It follows that $p+pq+q+q^2=93$. By parity, it must be $p=2$ and we get $q(3+q)=91$, i.e. no solution.

(f) $k_1=2,k_2=2$. Then, $n=p^2q^2$ and $d(n)=9$. It follows that $p+p^2+pq+p^2q+pq^2+q+q^2=90$. By parity, it must be $p=2$ and we get $q(7+3q)=84$, i.e. no solution.

(g) $k_1=3, k_2=2$. Then, $n=p^3q^2$ and $d(n)=12$. But $$p^3q+p^2q^2+pq^2+p^2q \geq 8\cdot3+4\cdot9+2\cdot9+4\cdot3>99-d(n)=87,$$ so there are no solutions.

(h) $k_1=4, k_2=2$. Then, $n=p^4q^2$ and $d(n)=15$. But $$p^4+p^3q^2 \geq 16+8\cdot9>99-d(n)=84,$$ so there are no solutions.

(i) $k_1=1, k_2=3$. Then, $n=pq^3$ and $d(n)=8$. It follows that $p+pq+pq^2+q+q^2+q^3=91$. By parity, it must be $p=2$ and we get $q(3+3q+q^2)=89$, i.e. no solution.

(j) $k_1=2,k_2=3$. Then, $n=p^2q^3$ and $d(n)=12$. But $$q^3+pq^3+pq^2 \geq 27+2\cdot27+2\cdot9>99-d(n)=87,$$ so there are no solutions.

(k) $k_1=3, k_2=3$. Then, $n=p^3q^3$ and $d(n)=16$. But $$p^2q^3 \geq 4\cdot27>99-d(n)=83,$$ so there are no solutions.

(l) $k_1=4, k_2=3$. Then, $n=p^4q^3$ and $d(n)=20$. But $$p^2q^3 \geq 4\cdot27>99-d(n)=79,$$ so there are no solutions.

(iii) $\omega(n)=3$. Then, $n=p^{k_1}q^{k_2}r^{k_3}$, where $p,q,r$ are prime numbers, $p<q<r$ and $k_1,k_2,k_3 \in \mathbb{N}^*$. Since $p \geq 2$, $q \geq 3$ and $r \geq 5$, if $n=p^3qr$, then $p^3q+p^3r+p^2qr \geq 8\cdot3+8\cdot5+4\cdot3\cdot5>97$, so $k_1 \leq 2$. If $n=pq^2r$, then $pq^2+q^2r+pqr+r \geq 2\cdot9+9\cdot5+2\cdot3\cdot5>97$, so $k_2=1$. If $n=pqr^2$, then $qr^2+pr^2 \geq 3\cdot 25+2\cdot25>97$, so $k_3=1$.  We have two cases.

(a) $k_1=k_2=k_3=1$. Then, $n=pqr$ and $d(n)=8$. It follows that $p+q+r+pq+pr+qr=91$. By parity, it must be $p=2$, which gives $3q+3r+qr=89$, i.e. $(q+3)(r+3)=98$. There are no solutions in this case.

(b) $k_2=2, k_2=k_3=1$. Then, $n=p^2qr$ and $d(n)=12$. It follows that $p+p^2+q+pq+p^2q+r+pr+p^2r+qr+pqr=87$. By parity, it must be $p=2$, which gives $7q+7r+3qr=81$. If $q \geq 5$, then $r \geq 7$ and $7q+7r+3qr>81$. So, $q=3$ and $4r=15$, i.e. no solution.

In conclusion, $n \in \{88,116\}$.

Mathematical Reflections 2016, Issue 4 - Problem J382

Problem:
Find all triples $(x,y,z)$ of real numbers with $x,y,z>1$ satisfying
$$\left(\dfrac{x}{2}+\dfrac{1}{x}-1\right)\left(\dfrac{y}{2}+\dfrac{1}{y}-1\right)\left(\dfrac{z}{2}+\dfrac{1}{z}-1\right)=\left(1-\dfrac{x}{yz}\right)\left(1-\dfrac{y}{zx}\right)\left(1-\dfrac{z}{xy}\right).$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
By the AM-GM Inequality, we have
$$\dfrac{x}{2}+\dfrac{1}{x}-1=\left(\dfrac{x}{2}+\dfrac{1}{2x}\right)+\dfrac{1}{2x}-1 \geq \dfrac{1}{2x} > 0$$
$$\dfrac{y}{2}+\dfrac{1}{y}-1=\left(\dfrac{y}{2}+\dfrac{1}{2y}\right)+\dfrac{1}{2y}-1 \geq \dfrac{1}{2y} > 0$$
$$\dfrac{z}{2}+\dfrac{1}{z}-1=\left(\dfrac{z}{2}+\dfrac{1}{2z}\right)+\dfrac{1}{2z}-1 \geq \dfrac{1}{2z} > 0.$$
Thus, $$\left(1-\dfrac{x}{yz}\right)\left(1-\dfrac{y}{zx}\right)\left(1-\dfrac{z}{xy}\right)>0.$$
Assume that two among these three factors are negative. Without loss of generality, $1-\dfrac{x}{yz}<0$ and $1-\dfrac{y}{zx}<0$. Then, $\dfrac{x}{yz}>1$ and $\dfrac{y}{zx}>1$, which gives $\dfrac{1}{z^2}>1$, contradiction. Hence,
$$1-\dfrac{x}{yz}>0, \qquad 1-\dfrac{y}{zx}>0, \qquad 1-\dfrac{z}{xy}>0.$$
We have $$\begin{array}{lll} \left(\dfrac{x}{2}+\dfrac{1}{x}-1\right)^2-\left(1-\dfrac{y}{zx}\right)\left(1-\dfrac{z}{xy}\right)&=&\dfrac{1}{x^2}+\dfrac{2}{x}\left(\dfrac{x}{2}-1\right)+\left(\dfrac{x}{2}-1\right)^2-\left[1-\dfrac{yz}{x}\left(\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)+\dfrac{1}{x^2}\right] \\ &=&\left(\dfrac{x}{2}-1\right)^2+\dfrac{yz}{x}\left(\dfrac{1}{y}-\dfrac{1}{z}\right)^2 \geq 0. \end{array}$$
The equality holds if and only if $x=2$ and $y=z$. Likewise,
$$\left(\dfrac{y}{2}+\dfrac{1}{y}-1\right)^2 \geq \left(1-\dfrac{z}{xy}\right)\left(1-\dfrac{x}{yz}\right)$$
$$\left(\dfrac{z}{2}+\dfrac{1}{z}-1\right)^2 \geq \left(1-\dfrac{x}{yz}\right)\left(1-\dfrac{y}{zx}\right)$$
Multiplying these three inequalities, we get that the left-hand side of the given equation is greater or equal than the right-hand side and the equality holds if and only if $x=y=z=2$. So, $(x,y,z)=(2,2,2)$.

Gazeta Matematica 4/2016, Problem S:L16.140

Problem:
Determine all prime numbers $p,q$ such that $$p^2+pq+q^2$$ is a perfect square.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that $p^2<p^2+pq+q^2<(p+q)^2$ for all primes $p,q$. Hence, $$p^2+pq+q^2=(p+k)^2,$$ where $k \in \{1,2,\ldots,q-1\}$, i.e. $$pq+q^2=2kp+k^2.$$ We get
$$p(2k-q)=(q-k)(q+k) \qquad (1)$$
Since the right-hand side is positive, it follows that $q<2k$. Moreover, since $p$ is prime, then $p \ | \ (q-k)$ or $p \ | \ (q+k)$.

(i) If $p \ | \ (q-k)$, then $q-k=ph$ for some $h \in \mathbb{N}$, so equation (1) becomes $$2k-q=h(k+q).$$ It follows that $(k+q) \ | \ (2k-q)$. Since $2k-q=2(k+q)-3q$, then $(k+q) \ | \ 3q$. If $q=3$, there is no solution. If $q \neq 3$, we have $k+q \in \{1,3,q,3q\}$. But $1<q<k+q<2q<3q$, so it must be $k+q=3$. Since $q$ is prime, this gives $q=2$ and $k=1$, but this would imply that $p \ | \ 1$, contradiction.

(ii) If $p \ | \ (q+k)$, then $q+k=ph$ for some $h \in \mathbb{N}$, so equation (1) becomes $$2k-q=h(q-k).$$ It follows that $(q-k) \ | \ (2k-q)$. Since $2k-q=2(k-q)+q$, then $(q-k) \ | \ q$. But $q-k<q$, so $q-k=1$. Equation (1) becomes $$p(q-2)=2q-1,$$ which implies $(q-2) \ | \ (2q-1)=[2(q-2)+3]$, i.e. $(q-2) \ | \ 3$. So, $q=3$ or $q=5$, which gives $p=5$ or $p=3$ respectively.

Therefore, $(p,q) \in \{(3,5),(5,3)\}$.

Note:
A variant of this problem can be the following:

Let $p,q$ be prime numbers such that $$p^2+pq+q^2$$ is a perfect square. Prove that $$p^2-pq+q^2$$ is prime.
\newline\newline
Indeed, $p$ and $q$ can be found as before and we get $p^2-pq+q^2=19$.

Gazeta Matematica 2/2016, Problem E:14971

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

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that $$\begin{array}{lll}(x+1)^2+(y+1)^2+xy(x+y+3)-1&=&x^2+y^2+2xy+2x+2y+1+xy(x+y+1)\\&=&(x+y+1)^2+xy(x+y+1)\\&=&(x+y+1)(x+1)(y+1), \end{array}$$ so the equation can be rewritten as $$(x+y+1)(x+1)(y+1)=1.$$ Therefore, at least one of the three factors is equal to $1$. We have the three systems of equations
$$\begin{array}{rll} x+y+1&=&1 \\ (x+1)(y+1)&=&1, \end{array} \qquad \begin{array}{rll} x+1&=&1 \\ (x+y+1)(y+1)&=&1, \end{array} \qquad \begin{array}{rll} y+1&=&1 \\ (x+y+1)(x+1)&=&1. \end{array}$$
Solving each system we get that $(x,y) \in \{(0,0),(-2,0),(0,-2)\}$.

Gazeta Matematica 2/2016, Problem E:14970

Problem:
Let $$\begin{array}{rcl}a_1&=&25 \\ a_n&=&\underbrace{22\ldots 2}_{n \textrm{ times}}\underbrace{44\ldots 4}_{n-1 \textrm{ times}}5, \qquad n \geq 2. \end{array}$$
Prove that $a_n$ can be written as the sum of two perfect squares for any integer $n \geq 1$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Clearly, $a_1=9+16=3^2+4^2$. Moreover, we have $$\begin{array}{rcl} a_n&=&\underbrace{22\ldots 2}_{n \textrm{ times}}\underbrace{44\ldots 4}_{n-1 \textrm{ times}}5 \\ &=& \underbrace{22\ldots 2}_{n \textrm{ times}}\cdot 10^n + \underbrace{44\ldots 4}_{n-1 \textrm{ times}}\cdot10+5 \\ &=& \dfrac{2\cdot(10^n-1)}{9}\cdot10^n+\dfrac{4(10^{n-1}-1)}{9}\cdot10+5 \\ &=& \dfrac{2\cdot10^{2n}-2\cdot10^n+4\cdot10^n+5}{9} \\ &=&\dfrac{10^{2n}-2\cdot10^n+1}{9}+\dfrac{10^{2n}+4\cdot10^n+4}{9} \\ &=& \left(\dfrac{10^n-1}{3}\right)^2+\left( \dfrac{10^n+2}{3} \right)^2  \end{array}.$$ Observe that $\dfrac{10^n-1}{3}$ and $\dfrac{10^n+2}{3}$ are integers since $10^n-1$ and $10^n+2$ are divisible by $3$ for any $n \geq 1$.

Mathematical Reflections 2016, Issue 3 - Problem O373

Problem:Let $n \geq 3$ be a natural number. On a $n \times n$ table we perform the following operation: choose a $(n-1) \times (n-1)$ square and add or subtract $1$ to all its entries. At the beginning all the entres in the table are $0$. Is it possible after a finite number of operations to obtain all the natural numbers from $1$ to $n^2$ in the table?

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
The answer is no. Observe that after each operation on a $(n-1) \times (n-1)$ square, the sum of all the elements in the $n \times n$ table is constant modulo $(n-1)^2$. Since at the beginning the table is filled with zeros, then after any finite number of operations the sum of the elements in the table is divisible by $(n-1)^2$. On the other hand, $$S=1+2+\ldots+n^2=\dfrac{n^2(n^2+1)}{2}.$$
If $n$ is even, then $n=2k$ where $k \geq 2$, and $$S=2k^2(4k^2+1).$$ Assume that $(2k-1)^2 \ | \ S$. Then $(2k-1)^2 \ | \ 2S=4k^2(4k^2+1)$. Since $(2k-1,2k)=1$, then $((2k-1)^2,4k^2)=1$. It follows that $(2k-1)^2 \ | \ (4k^2+1)$, i.e. $(2k-1)^2 \ | \ 4k$. This leads to $(2k-1)^2 \leq 4k$, which is false for $k \geq 2$.
If $n$ is odd, then $n=2k+1$, where $k \geq 1$ and $$S=(2k+1)^2(2k^2+2k+1).$$ Assume that $4k^2 \ | \ S$. Then $4k^2 \ | \ 2S=(2k+1)^2(4k^2+4k+2)$. Since $(2k,2k+1)=1$, then $(4k^2,(2k+1)^2)=1$. It follows that $4k^2 \ | \ (4k^2+4k+2)$, i.e. $4k^2 \ | \ (4k+2)$. This leads to $4k^2 \leq 4k+2$, which is false for $k \geq 2$. Moreover, if $k=1$, then $4 \nmid 6$.
In conclusion, $(n-1)^2$ doesn't divide $\dfrac{n^2(n^2+1)}{2}$ for any $n \geq 3$ and the conclusion follows.

Note:1) Observe that if $n=2$, it is possible to obtain all the natural numbers from $1$ to $4$ with the above operations, since it is possible to increase any $1 \times 1$ square by any quantity. This accords with the fact that $1$ divides $1+2+3+4=10$ trivially.
2) Li Zhou gave a very clever solution to the general problem in which the entries of the $(n-1) \times (n-1)$ square are not changed by the same quantity (for example to the first entry add $1$, and to the other entries add $-1$ and so on). See the original solution published on Mathematical Reflections 3/2016.

Mathematical Reflections 2016, Issue 3 - Problem U374

Problem:
Let $p$ and $q$ be complex numbers such that two of the zeros $a,b,c$ of the polynomial $x^3+3px^2+3qx+3pq=0$ are equal. Evaluate $a^2b+b^2c+c^2a$.

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

Solution:
Assume without loss of generality that $a=b$. Then, $a^2b+b^2c+c^2a=b^3+b^2c+bc^2$. By Vi\`ete's Formulas, we have $$\begin{array}{rcl} abc&=&-3pq \\ ab+bc+ca&=&3q \\ a+b+c&=&-3p. \end{array}$$ Since $a=b$, we have $$\begin{array}{rcl} b^2c&=&-3pq \\ b^2+2bc&=&3q \\ 2b+c&=&-3p. \end{array}$$ Multiplying side by side the last two equations, we get $$(b^2+2bc)(2b+c)=-9pq.$$ Since $-9pq=3(-3pq)=3b^2c$, we get $$(b^2+2bc)(2b+c)=3b^2c,$$ i.e. $$b^3+b^2c+bc^2=0.$$ It follows that $a^2b+b^2c+c^2a=0$.

Mathematical Reflections 2016, Issue 3 - Problem U373

Problem:
Prove the following inequality holds for all positive integers $n \geq 2$,
$$\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1+\dfrac{1}{1+2+\ldots+n}\right)<3.$$

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

Solution:
Since $1+2+\ldots+j=\dfrac{j(j+1)}{2}$ for any $j=2,\ldots,n$, we have $$\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1+\dfrac{1}{1+2+\ldots+n}\right)=\prod_{j=2}^n \left(1+\dfrac{2}{j(j+1)}\right).$$ Using the AM-GM Inequality, we have
$$\prod_{j=2}^n \left(1+\dfrac{2}{j(j+1)}\right) \leq \left(\dfrac{n-1+2\sum_{j=2}^n \frac{1}{j(j+1)}}{n-1}\right)^{n-1}=\left(1+\dfrac{1-\frac{2}{n+1}}{n-1}\right)^{n-1}=\left(1+\dfrac{1}{n+1}\right)^{n-1}.$$
By the Binomial Theorem, we have $$\left(1+\dfrac{1}{n+1}\right)^{n-1}=\sum_{k=0}^{n-1} {n-1 \choose k} \dfrac{1}{(n+1)^k}=\sum_{k=0}^{n-1} \dfrac{(n-1)!}{(n-1-k)!(n+1)^k}\cdot\dfrac{1}{k!}.$$ Since $\dfrac{(n-1)!}{(n-1-k)!(n+1)^k}<1$ and $\dfrac{1}{k!} \leq \dfrac{1}{2^{k-1}}$ for all $k \geq 2$, then
$$\left(1+\dfrac{1}{n+1}\right)^{n-1} \leq \sum_{k=0}^{n-1} \dfrac{1}{k!} < 1+1+\sum_{k=2}^{n-1} \dfrac{1}{2^{k-1}}<3,$$ which gives the desired conclusion.

Mathematical Reflections 2016, Issue 3 - Problem S374

Problem:
Let $a,b,c$ be positive real numbers. Prove that at least one of the numbers $$\dfrac{a+b}{a+b-c}, \qquad \dfrac{b+c}{b+c-a}, \qquad \dfrac{c+a}{c+a-b}$$ is not in the interval $(1,2)$.

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

Solution:
Assume by contradiction that all the given numbers are in the interval $(1,2)$. Then,
$$\dfrac{a+b}{a+b-c} > 1, \qquad \dfrac{b+c}{b+c-a} > 1, \qquad \dfrac{c+a}{c+a-b} > 1,$$ which gives
$$\dfrac{c}{a+b-c}>0, \qquad \dfrac{a}{b+c-a}>0, \qquad \dfrac{b}{c+a-b}>0.$$ Since $a,b,c$ are positive real numbers, then $a+b-c>0, b+c-a>0$ and $c+a-b>0$. Moreover,
$$\dfrac{a+b}{a+b-c} < 2, \qquad \dfrac{b+c}{b+c-a} < 2, \qquad \dfrac{c+a}{c+a-b} < 2,$$ which gives $$\dfrac{2c-a-b}{a+b-c}<0, \qquad \dfrac{2a-b-c}{b+c-a}<0, \qquad \dfrac{2b-c-a}{c+a-b}<0.$$ Since all the denominators are positive, then $2c-a-b<0$, $2a-b-c<0$, $2b-c-a<0$. Adding these three inequalities, we get $0<0$, contradiction. It follows that one of the three given numbers is not in the interval $(1,2)$.

Mathematical Reflections 2016, Issue 3 - Problem J375

Problem:
Solve in real numbers the equation $$\sqrt[3]{x}+\sqrt[3]{y}=\dfrac{1}{2}+\sqrt{x+y+\dfrac{1}{4}}$$

Proposed by Adrian Andreescu, Dallas, TX, USA

Solution:
Let $\sqrt[3]{x}=a$ and $\sqrt[3]{y}=b$. Then, $x=a^3$ and $y=b^3$ and the given equation becomes $$a+b-\dfrac{1}{2}=\sqrt{a^3+b^3+\dfrac{1}{4}}.$$ Squaring both sides, we have $$a^2+b^2+2ab-a-b=a^3+b^3,$$ i.e. $$(a+b)(a+b-1)=(a+b)(a^2-ab+b^2).$$ If $a+b=0$, then $x=-y$, but substituting into the original equation, we get $0=1$, contradiction. Then, $$a+b-1=a^2-ab+b^2,$$ i.e. $$a^2-a(b+1)+b^2-b+1=0.$$ Solving this equation for $a$, we obtain the discriminant $\Delta_a=(b+1)^2-4(b^2-b+1)=-3(b-1)^2 \leq 0$, so a real solutions exists if and only if $b-1=0$, i.e. $b=1$. Substituting this value, we get $(a-1)^2=0$, so $a=1$. We conclude that the given equation has only the real solution $(x,y)=(1,1)$.

Mathematical Reflections 2016, Issue 3 - Problem J373

Problem:
Let $a,b,c$ be real numbers greater than $-1$. Prove that $$(a^2+b^2+2)(b^2+c^2+2)(c^2+a^2+2) \geq (a+1)^2(b+1)^2(c+1)^2.$$

Proposed by Adrian Andreescu, Dallas, TX, USA

Solution:
By the AM-GM Inequality, we have $$a^2+b^2+2=(a^2+1)+(b^2+1) \geq 2\sqrt{(a^2+1)(b^2+1)}.$$ Likewise, $b^2+c^2+2 \geq 2\sqrt{(b^2+1)(c^2+1)}$ and $c^2+a^2+2 \geq 2\sqrt{(c^2+1)(a^2+1)}$, so
$$(a^2+b^2+2)(b^2+c^2+2)(c^2+a^2+2) \geq 8(a^2+1)(b^2+1)(c^2+1). \qquad                  (1)$$
By the Cauchy-Schwarz Inequality, we have $$2(a^2+1)=(1+1)(a^2+1) \geq (a+1)^2.$$
Likewise, $2(b^2+1) \geq (b+1)^2$ and $2(c^2+1) \geq (c+1)^2$, so
$$8(a^2+1)(b^2+1)(c^2+1) \geq (a+1)^2(b+1)^2(c+1)^2. \qquad                                (2)$$
By inequalities (1) and (2), we obtain the desired inequality.

Recreatii Matematice 1/2016, Problem VIII.201

Problem:
Prove that the number $2(n^4-n^2+1)$ is the sum of two perfect squares for all $n \in \mathbb{N}$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:We have $$\begin{array}{lll}2(n^4-n^2+1)&=&2n^2(n^2-1)+2\\&=&2n\cdot(n+1)n(n-1)+2\\&=&[(n+2)+(n-2)](n+1)n(n-1)+2\\&=&(n+2)(n+1)n(n-1)+1+(n+1)n(n-1)(n-2)+1 \\&=&[(n+2)(n-1)(n+1)n+1]+[(n+1)(n-2)n(n-1)+1]\\&=&[(n^2+n-2)(n^2+n)+1]+[(n^2-n-2)(n^2-n)+1]\\&=&(n^2+n-1)^2+(n^2-n-1)^2. \end{array}$$

Gazeta Matematica 12/2015, Problem S:E15.354

Problem:
Prove that the number $$a=2n^2+[\sqrt{4n^2+n}]+1$$ where $n$ is a nonzero natural number, can be written as the sum of two perfect squares.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Let us prove first that $[\sqrt{4n^2+n}]=2n$. Indeed, for all $n \in \mathbb{N}$ $$4n^2 \leq 4n^2+n < 4n^2+4n+1,$$ so $$2n \leq \sqrt{4n^2+n}<2n+1.$$ It follows that $[\sqrt{4n^2+n}]=2n$. So, $$a=2n^2+2n+1=n^2+(n+1)^2.$$  

Mathematical Reflections 2016, Issue 2 - Problem U372

Problem:
Let $\alpha,\beta>0$ be real numbers and let $f:\mathbb{R} \to \mathbb{R}$ be a continuous function such that $f(x) \neq 0$ for all $x$ in a neighborhood $U$ of $0$. Evaluate $$\lim_{x \to 0^+} \dfrac{\int_0^{\alpha x} t^{\alpha}f(t) \ dt}{\int_0^{\beta x} t^{\beta}f(t) \ dt}.$$

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

Solution:
Observe that $\lim_{x \to 0+} \int_0^{\alpha x} t^{\alpha}f(t) \ dt=\lim_{x \to 0^+} \int_0^{\beta x} t^{\beta}f(t) \ dt=0$. Let $g(t)=t^{\alpha}f(t)$ and $h(t)=t^{\beta} f(t)$. We have $$\dfrac{d}{dx} \int_0^{\alpha x} g(t) \ dt=g(\alpha x)\alpha=(\alpha x)^{\alpha}f(\alpha x)\alpha$$ and $$\dfrac{d}{dx} \int_0^{\beta x} h(t) \ dt=h(\beta x)\beta=(\beta x)^{\beta}f(\beta x)\beta.$$ Clearly, $\dfrac{d}{dx} \int_0^{\beta x} h(t) \ dt \neq 0$ for all $x$ in a neighborhood $U$ of $0$ ($x \neq 0$). We have $$\lim_{x \to 0^+}\dfrac{\dfrac{d}{dx} \int_0^{\alpha x} t^{\alpha}f(t) \ dt}{\dfrac{d}{dx} \int_0^{\beta x} t^{\beta}f(t) \ dt}=\begin{cases} 0 & \textrm{ if } \alpha>\beta \\ 1 & \textrm{ if } \alpha=\beta \\ +\infty & \textrm{ if } \alpha<\beta.  \end{cases}$$
By L'H\^{o}pital's Rule we conclude that
$$\lim_{x \to 0^+} \dfrac{\int_0^{\alpha x} t^{\alpha}f(t) \ dt}{\int_0^{\beta x} t^{\beta}f(t) \ dt}=\begin{cases} 0 & \textrm{ if } \alpha>\beta \\ 1 & \textrm{ if } \alpha=\beta \\ +\infty & \textrm{ if } \alpha<\beta.  \end{cases}$$

Mathematical Reflections 2016, Issue 2 - Problem U370

Problem:
Evaluate $$\lim_{x \to 0} \dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+3x}\cdot\ldots\cdot\sqrt[n]{1+nx}-1}{x}.$$

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

Solution:
Observe that for any $k=2,\ldots,n$ and for any $x$ such that $|x|<1$, we have $$\sqrt[k]{1+kx}=1+\dfrac{1}{k}kx+o(x^2)=1+x+o(x^2).$$ Hence,
$$\begin{array}{lll}\sqrt{1+2x}\cdot\sqrt[3]{1+3x}\cdot\ldots\cdot\sqrt[n]{1+nx}-1&=&(1+x+o(x^2))^{n-1}-1\\&=&(1+(n-1)x+o(x^2))-1\\&=&(n-1)x+o(x^2). \end{array}$$
Therefore, $$\lim_{x \to 0} \dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+3x}\cdot\ldots\cdot\sqrt[n]{1+nx}-1}{x}=n-1.$$

Mathematical Reflections 2016, Issue 2 - Problem U368

Problem:
Let $$x_n=\sqrt{2}+\sqrt[3]{\dfrac{3}{2}}+\ldots+\sqrt[n+1]{\dfrac{n+1}{n}}, \qquad n=1,2,3,\ldots$$
Evaluate $$\lim_{n \to \infty} \dfrac{x_n}{n}.$$

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

Solution:

Let $y_n=n$. Since $(y_n)_{n \geq 1}$ is strictly monotone and divergent sequence and
$$\lim_{n \to \infty} \dfrac{x_{n+1}-x_n}{y_{n+1}-y_n}=\lim_{n \to \infty} \left(1+\dfrac{1}{n+1}\right)^{\frac{1}{n+2}}=1,$$ then by the Stolz-Cesaro Theorem we have $$\lim_{n \to \infty} \dfrac{x_n}{n}=1.$$  

Mathematical Reflections 2016, Issue 2 - Problem S368

Problem:
Determine all the natural numbers $n$ such that $$\sigma(n)=n+55,$$ where $\sigma(n)$ denotes the sum of the divisors of $n$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Since $\sigma(n) \geq n+1$, then the sum of the proper divisors of $n$ is $54$. Denoting by $\omega(n)$ the number of distinct primes dividing $n$, observe that if $\omega(n) \geq 4$, then there exist four prime numbers $p<q<r<s$ that divide $n$. Since $p \geq 2$, $q \geq 3$, $r \geq 5$ and $s \geq 7$, then $qrs$ divides $n$ but $qrs \geq 3\cdot5\cdot7>54$, which contradicts the hypothesis. So, $\omega(n) \leq 3$. We have three cases.

(i) $\omega(n)=1$. Then, $n=p^k$, where $p$ is a prime number and $k \in \mathbb{N}^*$. Hence, $$1+p+\ldots+p^k=p^k+55 \implies p(1+p+\ldots+p^{k-2})=54.$$ It follows that $p \in \{2,3\}$. An easy check shows that there are no solutions.

(ii) $\omega(n)=2$. Then, $n=p^{k_1}q^{k_2}$, where $p,q$ are prime numbers, $p<q$ and $k_1,k_2 \in \mathbb{N}^*$. Since $p \geq 2$ and $q \geq 3$, if $n=p^4q$, then $p+p^2+p^3+p^4+p^3q+q \geq 57>54$, so $k_1 \leq 3$. If $n=pq^3$, then $pq^2+q+q^2+q^3 \geq 2\cdot9+39>54$, so $k_2 \leq 2$.  We have six cases.

(a) $k_1=1,k_2=1$. Then, $n=pq$ and $p+q=54$, which gives $$(p,q) \in \{(7,47),(11,43),(13,41),(17,37),(23,31)\}.$$

(b) $k_2=2, k_2=1$. Then, $n=p^2q$ and $p+p^2+q+pq=54$. Both $p$ and $q$ must be odd. Since $p+p^2<54$, then $p \in \{3,5\}$. An easy check shows that there are no solutions.

(c) $k_1=3, k_2=1$. Then, $n=p^3q$ and $p+p^2+p^3+q+pq+p^2q=54$. Both $p$ and $q$ must be odd. Since $5^3>54$, then $p=3$. An easy check shows that there are no solutions.

(d) $k_1=1, k_2=2$. Then, $n=pq^2$ and $p+pq+q+q^2=54$. Since $q+q^2<54$, then $q \in \{3,5\}$. An easy check shows that there are no solutions.

(e) $k_1=2,k_2=2$. Then, $n=p^2q^2$ and $p+p^2+pq+p^2q+pq^2+q+q^2=54$. At least one between $p$ and $q$ must be even, so $p=2$. Thus, we get $$(p,q)=(2,3).$$

(f) $k_1=3, k_2=2$. Then, $n=p^3q^2$, but $p^3q+p^2q^2 \geq 8\cdot3+4\cdot9>54$, so there are no solutions.

(iii) $\omega(n)=3$. Then, $n=p^{k_1}q^{k_2}r^{k_3}$, where $p,q,r$ are prime numbers, $p<q<r$ and $k_1,k_2,k_3 \in \mathbb{N}^*$. Since $p \geq 2$, $q \geq 3$ and $r \geq 5$, if $n=p^3qr$, then $p^3q+p^3r \geq 8\cdot3+8\cdot5>54$, so $k_1 \leq 2$. If $n=pq^2r$, then $pq^2+q^2r \geq 2\cdot9+9\cdot5>54$, so $k_2=1$. If $n=pqr^2$, then $qr^2 \geq 3\cdot 25>54$, so $k_3=1$.  We have two cases.

(a) $k_1=k_2=k_3=1$. Then, $n=pqr$ and $p+q+r+pq+pr+qr=54$. Clearly, $p \neq 2$. If $p \geq 5$, then $q \geq 7$ and $r \geq 11$, but $pr=55>54$, contradiction. So, $p=3$ and $4q+4r+qr=51$, i.e. $(q+4)(r+4)=67$, but $67$ is a prime number, so no solutions in this case.

(b) $k_2=2, k_2=k_3=1$. Then, $n=p^2qr$ and $p+p^2+q+pq+p^2q+r+pr+p^2r+qr+pqr=54$. Clearly, $p$ cannot be even. So, $p \geq 3$ but $p^2r \geq 9\cdot7>54$, contradiction.


In conclusion, $n \in \{36,329,473,533,629,713\}$.

Mathematical Reflections 2016, Issue 2 - Problem J371

Problem:
Prove that for all positive integers $n$, $${n+3 \choose 2} +6{n+4 \choose 4} + 90{n+5 \choose 6}$$ is the sum of two perfect cubes.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:

Let $$N=\dfrac{(n+5)!}{8(n-1)!}+\dfrac{(n+4)!}{4n!}+\dfrac{(n+3)!}{2(n+1)!}.$$
We have $$N=\dfrac{1}{8}[n(n+1)(n+2)(n+3)(n+4)(n+5)+2(n+1)(n+2)(n+3)(n+4)+4(n+2)(n+3)]$$
Set $x=n^2+5n$. Hence, $$\begin{array}{lll} N&=&\dfrac{1}{8}[x(x+4)(x+6)+2(x+4)(x+6)+4(x+6)]\\&=&\dfrac{1}{8}[x(x+4)(x+6)+2(x+4)(x+6)+4(x+4)+8]\\&=&\dfrac{1}{8}(x+4)(x(x+6)+2(x+6)+4)+1\\&=&\dfrac{1}{8}(x+4)^3+1\\&=&\left(\dfrac{n^2+5n+4}{2}\right)^3+1\\&=&\left(\dfrac{(n+1)(n+4)}{2}\right)^3+1,\end{array}$$ which is the sum of two perfect cubes since $(n+1)(n+4)$ is even for all $n \in \mathbb{N}^*$.

Mathematical Reflections 2016, Issue 2 - Problem J369

Problem:
Solve the equation $$\sqrt{1+\dfrac{1}{x+1}}+\dfrac{1}{\sqrt{x+1}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}.$$

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

Solution:
Clearly, $x>0$. The given equation can be written as $$\dfrac{\sqrt{x+2}}{\sqrt{x+1}}+\dfrac{1}{\sqrt{x+1}}=\dfrac{x+1}{\sqrt{x}},$$ i.e. $$\sqrt{x^2+2x}+\sqrt{x}=(x+1)\sqrt{x+1}.$$ The last equation can be written as $$\dfrac{\sqrt{x}}{\sqrt{x+2}-1}=\sqrt{x+1},$$ which gives $$\sqrt{x}+\sqrt{x+1}=\sqrt{(x+2)(x+1)}.$$ Squaring both sides and reordering, we get $$x^2+x+1=2\sqrt{x^2+x},$$ i.e. $$(\sqrt{x^2+x}-1)^2=0.$$ The last equation is equivalent to $x^2+x-1=0$, which solved for $x>0$ gives $x=\dfrac{-1+\sqrt{5}}{2}$.

Mathematical Reflections 2016, Issue 1 - Problem O361

Problem:
Determine the smallest natural number $n>2$ such that there exist $n$ consecutive integers whose sum of their squares is a perfect square.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that $$(-4)^2+(-3)^2+(-2)^2+(-1)^2+0^2+1^2+2^2+3^2+4^2+5^2+6^2=121=11^2,$$ so $n=11$ works. We prove that this is the smallest natural number greater than $2$ that satisfies the given condition.

(i) $n=3$. Let $m \in \mathbb{Z}$. We have $$(m-1)^2+m^2+(m+1)^2=3m^2+2 \equiv 2 \pmod{3},$$ so the sum of the squares of $3$ consecutive integers cannot be a perfect square.

(ii) $n=4$. Let $m \in \mathbb{Z}$. We have $$(m-1)^2+m^2+(m+1)^2+(m+2)^2=4m^2+4m+6 \equiv 2 \pmod{4},$$ so the sum of the squares of $4$ consecutive integers cannot be a perfect square.

(iii) $n=5$. Let $m \in \mathbb{Z}$. We have $$(m-2)^2+(m-1)^2+m^2+(m+1)^2+(m+2)^2=5(m^2+2),$$ but $m^2+2 \not \equiv 0 \pmod{5}$, so the sum of the squares of $5$ consecutive integers cannot be a perfect square.

(iv) $n=6$. Let $m \in \mathbb{Z}$. We have $$(m-2)^2+(m-1)^2+m^2+(m+1)^2+(m+2)^2+(m+3)^2=6m(m+1)+19.$$ Since $m(m+1)$ is even, then $6m(m+1) \equiv 0 \pmod{4}$ and $6m(m+1)+19 \equiv 3 \pmod{4}$. So the sum of the squares of $6$ consecutive integers cannot be a perfect square.

(v) $n=7$. Let $m \in \mathbb{Z}$. We have $$(m-3)^2+(m-2)^2+(m-1)^2+m^2+(m+1)^2+(m+2)^2+(m+3)^2=7(m^2+4),$$ but $m^2+4 \not \equiv 0 \pmod{7}$, so the sum of the squares of $7$ consecutive integers cannot be a perfect square.

(vi) $n=8$. Let $m \in \mathbb{Z}$. We have $$(m-3)^2+(m-2)^2+\ldots+(m+3)^2+(m+4)^2=4[2m(m+1)+11].$$ Since $m(m+1)$ is even, then $2m(m+1) \equiv 0 \pmod{4}$ and $2m(m+1)+11 \equiv 3 \pmod{4}$. So the sum of the squares of $8$ consecutive integers cannot be a perfect square.

(vii) $n=9$. Let $m \in \mathbb{Z}$. We have $$(m-4)^2+(m-3)^2+\ldots+(m+3)^2+(m+4)^2=3(3m^2+20),$$ but $3m^2+20 \not \equiv 0 \pmod{3}$, so the sum of the squares of $9$ consecutive integers cannot be a perfect square.

(viii) $n=10$. Let $m \in \mathbb{Z}$. We have $$(m-4)^2+(m-3)^2+\ldots+(m+4)^2+(m+5)^2=5[2m(m+1)+17],$$ but $2m(m+1)+17 \not \equiv 0 \pmod{5}$, so the sum of the squares of $10$ consecutive integers cannot be a perfect square.

The conclusion follows.

Note:
The solution is not unique. For example, $$18^2+19^2+20^2+21^2+22^2+23^2+24^2+25^2+26^2+27^2+28^2=77^2.$$
It's easy to see why. We must find $m \in \mathbb{Z}$ such that $$(m-5)^2+(m-4)^2+\ldots+(m+4)^2+(m+5)^2=11(m^2+10)$$ is a perfect square. So, $m \equiv \pm 1 \pmod{11}$ and one can check if $11(m^2+10)$ is a perfect square.

Conjecture:An easy case by case analysis shows that there are no solutions for $11<n \leq 22$. For $n=23$ there is a solution given by $$17^2+18^2+19^2+\ldots+27^2+28^2+29^2=138^2.$$ (Observe that $(m-11)^2+\ldots+(m+11)^2=23(m^2+44)$ and it must be $m \equiv \pm 5 \pmod{23}$ and $m \equiv 0 \pmod{2}$)

If $n>1$ is a natural number $n$ such that there exist $n$ consecutive integers whose sum of their squares is a perfect square, then $n$ is prime.

Mathematical Reflections 2016, Issue 1 - Problem U366

Problem:
If $f:[0,1] \to \mathbb{R}$ is a convex and integrable function with $f(0)=0$, prove that
$$\int_{0}^1 f(x) \ dx \geq 4\int_{0}^{\frac{1}{2}} f(x) \ dx.$$

Proposed by Florin Stanescu, Gaesti, Romania

Solution:
Since $f$ is a convex function, we have $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle \int_{0}^1 f(x) \ dx= \displaystyle \dfrac{1}{2}\left[\int_{0}^1 f(x) \ dx+\int_{0}^1 f(1-x) \ dx \right]&=&\displaystyle \int_{0}^1 \dfrac{f(x)+f(1-x)}{2} \ dx \\ &\geq& \displaystyle \int_{0}^1 f\left(\dfrac{x+(1-x)}{2}\right) \ dx\\&=&f\left(\dfrac{1}{2}\right). \end{array}$$
On the other hand, since $f$ is convex and $f(0)=0$, we have $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle f\left(\dfrac{1}{2}\right)=2\cdot\dfrac{f(0)+f\left(\frac{1}{2}\right)}{2}&=& \displaystyle 2\int_{0}^1 \left[(1-x)f(0)+xf\left(\dfrac{1}{2}\right)\right] \ dx \\ & \geq & \displaystyle 2\int_{0}^1 f\left((1-x)\cdot0+x\cdot\dfrac{1}{2}\right) \ dx \\&=& \displaystyle 4\int_{0}^{\frac{1}{2}} f(x) \ dx, \end{array}$$ and the conclusion follows.

Mathematical Reflections 2016, Issue 1 - Problem U365

Problem:
Let $n$ be a positive integer. Evaluate

(a) $\displaystyle \int_{0}^n e^{\lfloor x \rfloor} \ dx$

(b) $\displaystyle \int_{0}^n \lfloor e^x \rfloor \ dx$.

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

Solution:
(a) If $k \leq x < k+1$, where $k \in \mathbb{Z}$, then $\lfloor x \rfloor =k$. So, $$\displaystyle \int_{0}^n e^{\lfloor x \rfloor} \ dx=\sum_{k=0}^{n-1} \int_{k}^{k+1} e^{\lfloor x \rfloor} dx=\sum_{k=0}^{n-1} e^k=\dfrac{e^n-1}{e-1}.$$

(b) If $k \leq e^x < k+1$, where $k \in \mathbb{N}^*$, then $\lfloor e^x \rfloor=k$. This implies that if $\log k \leq x < \log (k+1)$, then $\lfloor e^x \rfloor=k$. Let $m$ be the greatest natural number such that $\log m \leq n$.
So, $$\renewcommand{\arraystretch}{2}\begin{array}{lll}\displaystyle \int_{0}^n \lfloor e^x \rfloor \ dx&=& \displaystyle \sum_{k=1}^{m-1} \int_{\log k}^{\log k+1} k \ dx+\int_{\log m}^n m \ dx\\&=&\displaystyle \sum_{k=1}^{m-1} [(k+1)\log(k+1)-k\log k]-\sum_{k=1}^{m-1} \log(k+1)+m(n-\log m)\\&=&\displaystyle m\log m-\log m!+mn-m\log m\\&=&mn-\log m!\\&=&\lfloor e^n \rfloor n-\log (\lfloor e^n \rfloor!). \end{array}$$

Mathematical Reflections 2016, Issue 1 - Problem U364

Problem:
Evaluate $$\int \dfrac{5x^2-x-4}{x^5+x^4+1}\ dx.$$

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

Solution:
Observe that $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \dfrac{5x^2-x-4}{x^5+x^4+1}&=&\dfrac{5x^2-x-4}{(x^3-x+1)(x^2+x+1)}\\&=&-\dfrac{3(x+1)}{x^2+x+1}+\dfrac{3x^2-1}{x^3-x+1}\\&=&-\dfrac{3}{2}\cdot\dfrac{2x+1}{x^2+x+1}+\dfrac{3x^2-1}{x^3-x+1}-\dfrac{3}{2}\cdot\dfrac{1}{x^2+x+1} \\ &=&-\dfrac{3}{2}\cdot\dfrac{2x+1}{x^2+x+1}+\dfrac{3x^2-1}{x^3-x+1}-\sqrt{3}\cdot\dfrac{2/\sqrt{3}}{\left(\frac{2x+1}{\sqrt{3}}\right)^2+1}. \end{array}$$
So, $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle \int \dfrac{5x^2-x-4}{x^5+x^4+1}\ dx&=&\displaystyle -\dfrac{3}{2}\int \dfrac{2x+1}{x^2+x+1} \ dx+\int \dfrac{3x^2-1}{x^3-x+1} \ dx-\sqrt{3}\int \dfrac{2/\sqrt{3}}{\left(\frac{2x+1}{\sqrt{3}}\right)^2+1} \ dx \\ &=&-\dfrac{3}{2}\log(x^2+x+1)+\log|x^3-x+1|-\sqrt{3}\arctan \dfrac{2x+1}{\sqrt{3}}+C. \end{array}$$

Mathematical Reflections 2016, Issue 1 - Problem S365

Problem:
Let $$a_k=\dfrac{(k^2+1)^2}{k^4+4}, \qquad k=1,2,3,\ldots.$$ Prove that for every positive integer $n$, $$a_1^na_2^{n-1}a_3^{n-2}\cdot \ldots \cdot a_n=\dfrac{2^{n+1}}{n^2+2n+2}.$$

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

Solution:
For any $k \in \mathbb{N}^*$, we have $$\renewcommand{\arraystretch}{2} \begin{array}{lll} a_k&=&\dfrac{(k^2+1)^2}{(k^2+2)^2-4k^2}\\&=&\dfrac{(k^2+1)^2}{(k^2-2k+2)(k^2+2k+2)}\\&=&\dfrac{k^2+1}{(k-1)^2+1}\cdot\dfrac{k^2+1}{(k+1)^2+1}. \end{array}$$
Let $P_j=\displaystyle \prod_{k=1}^j a_k$, where $j \in \mathbb{N}^*$. Then, $$P_j=\prod_{k=1}^j \dfrac{k^2+1}{(k-1)^2+1} \prod_{k=1}^j \dfrac{k^2+1}{(k+1)^2+1}=(j^2+1)\cdot\dfrac{2}{(j+1)^2+1}.$$ Therefore, we have
$$\renewcommand{\arraystretch}{2} \begin{array}{lll}\displaystyle a_1^na_2^{n-1}a_3^{n-2}\cdot \ldots \cdot a_n&=& \displaystyle \prod_{j=1}^n P_j\\ &=&\displaystyle\prod_{j=1}^n \dfrac{2(j^2+1)}{(j+1)^2+1}\\ &=& \displaystyle 2^n \prod_{j=1}^n \dfrac{j^2+1}{(j+1)^2+1}\\ &=&\displaystyle 2^n\cdot \dfrac{2}{(n+1)^2+1}\\&=&\displaystyle \dfrac{2^{n+1}}{n^2+2n+2}. \end{array}$$

Mathematical Reflections 2016, Issue 1 - Problem S362

Problem:
Let $0 < a,b,c,d \leq 1$. Prove that $$\dfrac{1}{a+b+c+d} \geq \dfrac{1}{4}+\dfrac{64}{27}(1-a)(1-b)(1-c)(1-d).$$

Proposed by An Zhen-ping, Xianyang Normal University, China

Solution:
The given inequality can be written as $$\dfrac{4-a-b-c-d}{4(a+b+c+d)} \geq \dfrac{64}{27}(1-a)(1-b)(1-c)(1-d).$$
By the AM-GM Inequality, we have $$\dfrac{64}{27}(1-a)(1-b)(1-c)(1-d) \leq \dfrac{64}{27}\left(\dfrac{4-a-b-c-d}{4}\right)^4=\dfrac{1}{4}\cdot\dfrac{1}{27}(4-a-b-c-d)^4.$$
So, we have to prove that $$\dfrac{1}{4}\cdot\dfrac{1}{27}(4-a-b-c-d)^4 \leq \dfrac{4-a-b-c-d}{4(a+b+c+d)},$$ i.e. $$\dfrac{1}{27}(4-a-b-c-d)^3 \leq \dfrac{1}{a+b+c+d}.$$ Let $x=a+b+c+d$. We have to prove that $x(4-x)^3 \leq 27$ for any $x \in (0,4]$, but this is true because $$27-x(4-x)^3=(x-1)^2(x^2-10x+27) \geq 0$$ for all $x \in (0,4]$. The equality holds if and only if $x=1$ and $1-a=1-b=1-c=1-d$, i.e. if and only if $a=b=c=d=\dfrac{1}{4}$.

Mathematical Reflections 2016, Issue 1 - Problem S361

Problem:
Find all integers $n$ for which there are integers $a$ and $b$ such that $(a+bi)^4=n+2016i$.

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

Solution:
We have $(a+bi)^4=(a^4-6a^2b^2+b^4)+4ab(a^2-b^2)i$, so it must be
$$\begin{array}{rcl} a^4-6a^2b^2+b^4&=&n \\ ab(a^2-b^2)&=&504. \end{array}$$
If $(a,b)$ is a solution to this system of equations, then also $(-a,-b)$, $(b,-a)$ and $(-b,a)$ are solutions, so we can assume $a^2-b^2>0$ and so $ab>0$. If $a^2-b^2 \equiv 0 \pmod{4}$, then both $a$ and $b$ are even and $ab(a^2-b^2)$ is divisible by $16$, contradiction.  Moreover, since $(a^2-b^2) \ | \ 504$ and $a^2-b^2 \not \equiv 2 \pmod{4}$ for all integers $a,b$, then $(a^2-b^2) \in \{1,3,7,9,21,63\}$. If $a^2-b^2=1$, then $a=\pm 1$ and $b=0$, which gives $ab=0$, contradiction. If $a^2-b^2=3$, then $a=\pm 2$ and $b=\pm 1$, which gives $ab=2$, contradiction. If $a^2-b^2=7$, then $a=\pm 4$ and $b=\pm 3$, which gives $ab=12$, contradiction. If $a^2-b^2=9$, then $a=\pm 5$ and $b=\pm 4$, which gives $ab=20$, contradiction. If $a^2-b^2=21$, then $a=\pm 11$ and $b=\pm 10$ or $a=\pm 5$ and $b=\pm 2$, which gives $ab \in \{110,10\}$, contradiction. If $a^2-b^2=63$, then $a=\pm 32$ and $b=\pm 31$ or $a=\pm 12$ and $b=\pm 9$ or $a=\pm 8$ and $b=\pm 1$, which gives $ab \in \{992,108,8\}$. We conclude that $a^2-b^2=63$ and $ab=8$, so $$a^4-6a^2b^2+b^4=(a^2-b^2)^2-4(ab)^2=3713,$$ which yields $n=3713$.

Mathematical Reflections 2016, Issue 1 - Problem J363

Problem:
Solve in integers the system of equations $$\begin{array}{rcl} x^2+y^2-z(x+y)&=&10 \\ y^2+z^2-x(y+z)&=&6 \\ z^2+x^2-y(z+x)&=&-2. \end{array}$$

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

Solution:

Observe that if $(x_0,y_0,z_0)$ is a solution to the given system, then also $(-x_0,-y_0,-z_0)$ is a solution. So, if there exists a solution, we can assume without loss of generality that $x_0+y_0+z_0 \geq 0$.
Adding the first equation to the third equation and subtracting the second, we get $$x^2-yz=1.$$ Adding the first equation to the second equation and subtracting the third, we get $$y^2-zx=9.$$ Adding the second equation to the third equation and subtracting the first, we get $$z^2-xy=-3.$$ So, the given system becomes:
$$\begin{array}{rcl} x^2-yz&=&1 \\ y^2-zx&=&9 \\ z^2-xy&=&-3. \end{array}$$
Subctracting the first two equations, then the last two equations and then the first equation and the third equation, we obtain the system of equations
$$\begin{array}{rcl} (x-y)(x+y+z)&=&-8 \\ (y-z)(x+y+z)&=&12 \\ (z-x)(x+y+z)&=&-4. \end{array}$$
Since by our assumption we have $x+y+z \geq 0$, we obtain that $x+y+z \in \{1,2,4\}$. We have three cases.

(i) $x+y+z=1$. Hence, we have $x-y=-8$ and $z-x=-4$, which gives $y=x+8$ and $z=x-4$. Therefore, $x+(x+8)+(x-4)=1$, which gives $x=-1,y=7,z=-5$.
(ii) $x+y+z=2$. Hence, we have $x-y=-4$ and $z-x=-2$, which gives $y=x+4$ and $z=x-2$. Therefore, $x+(x+4)+(x-2)=2$, which gives $x=0,y=4,z=-2$.
(iii) $x+y+z=4$. Hence, we have $x-y=-2$ and $z-x=-1$, which gives $y=x+2$ and $z=x-1$. Therefore, $x+(x+2)+(x-1)=4$, which gives $x=1,y=3,z=0$.

An easy check shows that the only solutions are $$(x,y,z) \in \{(1,3,0),(-1,-3,0)\}.$$

Mathematical Reflections 2016, Issue 1 - Problem J362

Problem:
Let $a,b,c,d$ be real numbers such that $abcd=1$. Prove that the following inequality holds: $$ab+bc+cd+da \leq \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1}{d^2}.$$

Proposed by Mircea Becheanu, University of Bucharest, Romania

Solution:
Without loss of generality, assume that $a \leq b \leq c \leq d$. Then, $\dfrac{1}{a} \geq \dfrac{1}{b} \geq \dfrac{1}{c} \geq \dfrac{1}{d}$. By the Rearrangement Inequality, $$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1}{d^2} \geq \dfrac{1}{cd}+\dfrac{1}{da}+\dfrac{1}{ab}+\dfrac{1}{bc}=ab+bc+cd+da.$$

Mathematical Reflections 2016, Issue 1 - Problem J361

Problem:
Solve in positive integers the equation $$\dfrac{x^2-y}{8x-y^2}=\dfrac{y}{x}.$$

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

Solution:

The given equation can be retwritten as $$x^3+y^3+27-9xy=27.$$ Using the well-known identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca),$$ we obtain $$(x+y+3)(x^2+y^2+9-xy-3x-3y)=27.$$ Since $x,y$ are positive integers, then $x+y+3 \geq 5$ and we have two cases.

(i) $x+y+3=9$ and $x^2+y^2+9-xy-3(x+y)=3$. From the first equation we have $x+y=6$, which gives $x^2+y^2=36-2xy$. Substituting these values into the second equation, we have $(36-2xy)+9-xy-18=3$, which gives $xy=8$. So, we obtain the system of equations
$$\begin{array}{rcl} x+y&=&6 \\ xy&=&8, \end{array}$$ which gives $(x,y) \in \{(2,4),(4,2)\}$. An easy check shows that the only solution to the given equation is $(x,y)=(4,2)$.

(ii) $x+y+3=27$ and $x^2+y^2+9-xy-3x-3y=1$.  From the first equation we have $x+y=24$, which gives $x^2+y^2=576-2xy$. Substituting these values into the second equation, we have $(576-2xy)+9-xy-72=3$, which gives $xy=170$. So, we obtain the system of equations $$\begin{array}{rcl} x+y&=&24 \\ xy&=&170, \end{array}$$ which yields no solutions in positive integers.

In conclusion, $(x,y)=(4,2)$. 

Gazeta Matematica 11/2015, Problem E:14911

Problem:
In a school, each student knows exactly $2n+1$ other students, where $n \in \mathbb{N}^*$.
Prove that the number of students in the school is even. (Suppose that "`knowing"' is a symmetrical relation).

Proposed by Alessandro Ventullo, Milan, Italy

Solution:

Suppose that two students know each other if and only if they shake their hands. Let $N$ be the total number of the handshakings and let $m$ be the total number of students. Since each student shakes his hand with $2n+1$ other students, there are $m(2n+1)$ handshakings. Since in each handshake are involved two students, we have counted each handshaking twice (one time when student $A$ shakes his hand with student $B$ and one time when student $B$ shakes his hand with student $A$). Therefore, the total number of handshakings is $N=\dfrac{m(2n+1)}{2}$. Since $N$ is a natural number, then $m$ must be even. 

Gazeta Matematica 10/2015, Problem 27137

Problem:
Evaluate $$\lim_{y \to 0} \int_{y}^1 \left(\arctan x \arctan \dfrac{1}{x}+\dfrac{\log(1+x^2)}{1+x^2}\right) \ dx.$$

Proposed by Alessandro Ventullo, Milan, Italy

 

Solution:

Observe that $$\arctan x + \arctan \dfrac{1}{x}=\dfrac{\pi}{2}$$ for all $x>0$. Setting $a=\arctan x, b=\arctan \dfrac{1}{x}, c=-\dfrac{\pi}{2}$, we have $a+b+c=0$, so $$a^3+b^3+c^3=3abc.$$ This means that
$$-3\arctan x \cdot \arctan \dfrac{1}{x} \cdot \dfrac{\pi}{2}=\arctan^3 x + \arctan^3 \dfrac{1}{x}-\dfrac{\pi^3}{8}.$$
Hence
$$\int_0^1 \arctan x \arctan \dfrac{1}{x} \ dx=-\dfrac{2}{3\pi}\int_0^1 \left( \arctan^3 x + \arctan^3 \dfrac{1}{x}-\dfrac{\pi^3}{8}\right) \ dx.                  (1)$$
Integrating by part, we obtain
$$\int \arctan^3 x \ dx=x \arctan^3 x - \dfrac{3}{2}\log(1+x^2)\arctan^2 x+3 \int \dfrac{\log (1+x^2) \arctan x}{1+x^2} \ dx.$$
and
$$\int \arctan^3 \dfrac{1}{x} \ dx=x \arctan^3 \dfrac{1}{x} + \dfrac{3}{2}\log(1+x^2)\arctan^2 \dfrac{1}{x}+3 \int \dfrac{\log (1+x^2) \arctan 1/x}{1+x^2} \ dx.$$
Summing up and integrating from $x=0$ to $x=1$ and using the fact that $$\arctan x+\arctan \dfrac{1}{x}=\dfrac{\pi}{2},$$ we get
$$\int_0^1 \left( \arctan^3 x + \arctan^3 \dfrac{1}{x} \right) \ dx=\dfrac{\pi^3}{32}+\dfrac{3\pi}{2} \int_0^1 \dfrac{\log (1+x^2)}{1+x^2} \ dx.                 (2)$$
Substituting (1) into (2), we get
$$\int_{0}^1 \left(\arctan x \arctan \dfrac{1}{x}+\dfrac{\log(1+x^2)}{1+x^2}\right)=\dfrac{\pi^2}{16}.$$  

Gazeta Matematica 6-7-8/2015, Problem E:14869

Problem:
Find all primes $p$ such that the equation
$$x^2(y+z)+y^2(z+x)+z^2(x+y)+3xyz=p^2$$
has positive integers solutions.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that $$x^2(y+z)+y^2(z+x)+z^2(x+y)+3xyz=(x+y+z)(xy+yz+zx),$$
so the equation can be rewritten as $$(x+y+z)(xy+yz+zx)=p^2.$$
Since $x,y,z$ must be positive integers, then $x+y+z>1$ and $xy+yz+zx>1$. This gives $x+y+z=xy+yz+zx=p$. Therefore, $$x(y-1)+y(z-1)+z(x-1)=0.$$ Note that each term on the left hand side is non-negative, so $$x(y-1)=y(z-1)=z(x-1)=0,$$ which means that $x=y=z=1$. Thus, $p=3$ is the only prime which satisfies the required property.