Let $m=dx$ and $n=dy$, where $d=\gcd(m,n)$, clearly $\gcd(x^5-y^5,xy)=1$, thus $d^3(x^5-y^5)=16xy$ implies that $x^5-y^5 | 16$, but because the difference between two consecutive powers of $5$ is either $1$ or is bigger than $30$, this easily implies that either $x$ or $y$ is $0$ or $xy=1$ or $xy=-1$, WLOG assume that $x=0$, thus $m=0$ and $n=0$ as well. Clearly $xy\neq 1$, if $xy=-1$ then $d=\pm 8$, and as pointed @below(thanks btw), this yields that $(m,n)=\{(2,-2), (-2,2)\}$ is also a solution.

rmtf1111 wrote: Let $m=dx$ and $n=dy$, where $d=\gcd(m,n)$, clearly $\gcd(x^5-y^5,xy)=1$, thus $d^3(x^5-y^5)=16xy$ implies that $x^5-y^5 | 16$, but because the difference between two consecutive powers of $5$ is either $1$ or is bigger than $30$, this easily implies that either $x$ or $y$ is $0$, WLOG assume that $x=0$, thus $m=0$ and $n=0$ as well. $m=-2, n=2$ is also a solution.

clearly $\gcd(x^5-y^5,xy)=1$.. Why?

IMO2019 wrote: clearly $\gcd(x^5-y^5,xy)=1$.. Why? $gcd(x, y) =1$

IMO2019 wrote: clearly $\gcd(x^5-y^5,xy)=1$.. Why? Since $(x^5,y^5) = 1$, then $(x^5 - y^5, x^5y^5) = 1$

Can you post the other problems, as well?

Very similar to BMO 2017 P1. That helped me to solve this problem on the competition.

I have a different solution involving bounding

x,y!=0; d=(m,n) m=dx,n=dy, d^3(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)=16xy (x,y)=1 (xy,x-y)=1 (xy,x^4+x^3y+x^2y^2+xy^3+y^4)=1 implying x^5-y^5|16 but x^4+x^3y+x^2y^2+xy^3+y^4 is odd (x,y are not both even ) so x^4+x^3y+x^2y^2+xy^3+y^4=1 and only sols is x=1 y=-1 and x=-1 y=1.Then m ,n are 2,-2; -2,2;0,0.

FISHMJ25 wrote: $x,y!=0$; $d=(m,n)$$m=dx$,$n=dy$, $d^3(x-y)$$(x^4+x^3y+x^2y^2+xy^3+y^4)=16xy$$(x,y)=1$$(xy,x-y)=1$$ (xy,x^4+x^3y+x^2y^2+xy^3+y^4)=1$implying $x^5-y^5|16$ but $x^4+x^3y+x^2y^2+xy^3+y^4$ is odd (x,y are not both even ) so $x^4+x^3y+x^2y^2+xy^3+y^4=1$ and only sols is $x=1$$y=-1$and $x=-1$$ y=1$.Then m ,n are $2,-2; -2,2;0,0$

Purple_Planet wrote:

In case 4 you didn't consider one of m' and n being less than 3 and the other being greater or equal.

$m^5-n^5=16mn$ Case 1: $m,n>0$ Let $(m,n)=d$, so that $m=dx, n=dy$. note that this implies $d^3(x^5-y^5)=16xy$. Then $d^3y^5$ is divisible by $x$ which is impossible unless $x=1$. Similarly $y=1$. However putting in the values we see that there's no solution in this case. Case 2: $m>0, n<0$ Let $n=-a, a >0$. Notice that $m^5+a^5=-16am$. Since the $LHS > 0$, there is no solution in this case. Case 3 : $m<0, n>0$ Let $m=-a, a>0$. Notice that $a^5+n^5=16an$. Let $(a,n)=d, a=dx, n=dy$. Notice that $d^3(x^5+y^5)=16xy$. This implies $d^3x^5$ is divisible by $y$, but because $(d^3x^5,y)=1, y=1$. Similarly $x=1$. Put in the values to get $d^3 \cdot 2 = 16 \implies d=2$. So, $m=-2, n=2$ Case 4: $m<0, n<0$ Notice that in this case the signs of the LHS and RHS don't match; no solutions here. Case 5: WLOG $m=0$ Then we get $n=0$ as well. So $(m,n)=(0,0)$ is a solution as well Combining these cases we get $(m,n)=(-2,2),(0,0)$

Hamroldt wrote: Then $d^3y^5$ is divisible by $x$ which is impossible unless $x=1$. Similarly $y=1$. how is this true, like $gcd(x,y)=1$ but it is possible that $x \mid d^3$, egs $(m,n) = (8,10)$, here $d = 2, x=4, y=5$ but $4 \mid 2^3$

First, if either $m = 0$ or $n = 0$ we must have $m = n = 0$, so we can assume $m$ and $n$ are nonzero from now on. Consider any odd prime $p$. If $v_p{(m)} = v_p{(n)} = e$, we get $v_p{(\text{LHS})} \ge 5e$ but $v_p{(\text{LHS})} = 2e$, absurd unless $e = 0$. Hence, $v_p{(m)}\neq v_p{(n)}$, so $v_p{(\text{LHS})} = 5\min{(v_p{(m)},v_p{(n)})}$ and $v_p{(\text{RHS})} = v_p{(m)} + v_p{(n)}$. This means either $v_p{(m)} = 4v_p{(n)}$ or $v_p{(n)} = 4v_p{(m)}$. Now write $m = da$ and $n = db$. The equation becomes $a^5 - b^5 = \pm 2^k$ for some integer $k$. But remember that $a$ and $b$ are relatively prime, so at most one of them is odd. This means either $a^5 - b^5 = \pm 1$, obviously impossible since $a$ and $b$ are nonzero, or $a$ and $b$ are both odd. But looking back at the original equation, since $4 - 3v_2{(d)}$ is positive, we must have $v_2{(d)} = 0$ or $v_2{(d)} = 1$, and hence $a^5 - b^5 = \pm 16$ or $a^5 - b^5 = \pm 2$, respectively. It is easy to see that the former equation yields no solutions and the latter yields only $(a,b) = (1, -1)$ and $(a,b) = (-1,1)$, which translate to $(m,n) = (2,-2)$ and $(m,n) = (-2,2)$, respectively. Only the latter is a solution, so we are done.

Solved with jacoporizzo. Clearly, $m\neq n$ and neither of the two equal zero unless both are zero. If there is a solution $(m,n)$ with $m$ and $n$ both positive or both negative, then note that $(-n,-m)$ is also a solution. Thus, it suffices to resolve the cases with $m$ and $n$ both positive or with $m$ negative and $n$ positive. If $m$ and $n$ are both positive, clearly $m>n$, so using $m^5-n^5= m^5-(m-1)^5\geq m^4$, we have \[m^4\leq 16mn\implies m^3\leq 16n\implies (m,n)=(2,1), (3,2).\]It is easy to verify that these are not solutions. If $m$ is negative and $n$ is positive, then substituting $m'=-m$ yields $$m'^5+n^5 = 16 m'n.$$By AM-GM, this implies that \[2m'^{5/2}\cdot n^{5/2}\leq 16m'n\implies m'n\leq 4.\]Indeed, checking $(m',n)=(2,2), (1,3), (1,4), (1,2), (1,1)$ yields that only $(m',n)=(2,2)$ works. Thus, the only solutions are $(m,n)=(0,0)$ and $(-2,2)$.

first of all if $m=n$ then we have trivially $(0,0)$ to be a valid pair, now we consider $m \neq n$ then: Case 1:- $m=a, n=b$ where $a,b>0$ now clearly we have $a^{5}-b^{5}=16ab \implies a^{5}-b^{5}=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4) =16ab$ since $a,b \in \mathbb{Z}^{+}$ we have $a^4+a^3b+a^2b^2+ab^3+b^4 \leqslant 16ab$ and by AM-GM we have $a^4+a^3b+a^2b^2+ab^3+b^4 \geqslant 5a^2b^2$ so we have $5a^2b^2 \leqslant 16ab \implies ab \leqslant 3$ which forces $(a,b):(1,1),(2,1),(3,1)$ , which we can see that they don't work. Case 2:- $m=-a , n=b$ where $ a,b>0$ we have $-a^5-b^5=-16ab \implies a^5+b^5=16ab$ also by AM-GM we have $\frac{a^5+b^5}{2} \geqslant a^2b^2 \sqrt{ab}\implies 4 \leqslant ab $ which forces $(a,b):(1,1),(2,1),(2,2),(3,1),(4,1)$ and their permutations , quick parity checking and plugging these we get only $(2,2)$ works which gives $m=-2, n=2$ Case 3:- $m=a,n=-b$ where $ a,b>0$ in this case we have $a^5+b^5=-16ab$ which has no solution over integers Case 4:- $m=-a,n=-b$ where $ a,b>0$ this gives $-a^5-b^5=16ab$ , which yet again has no solution over integers hence we have only possible pairs to be working are given by this set : $\boxed{\{(m,n) :(0,0),(-2,2)\}}$ $\blacksquare$