Find all the integer pairs $(x, y)$ which satisfy the equation $x^5-y^5=16xy$

First of all,if $x=y$ then $x=y=0$ which does work. So,now assume $x \not =y$. Again if one of them is $0$, other one has to be also. So,from now on also assume that none of them is $0$

$\textbf{Case 1:}$ $x,y$ both are positive. Then $(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)=16xy$

since obviously $x >y$,if $x \ge 3$ we have $x^4+x^3y \ge 9x^2+9xy \ge 9xy+9xy=18xy$.So, $x \le 2$So,only a few case to check.

$\textbf{Case 2:}$ Both are negative gives the same equation with $x,y$ swapped.

$\textbf{Case 3:}$ $x$ negative but $y$ positive would give $x^5+y^5=16xy$ by substituting $x=-x$ to make things easier to work with. Here a simple AM-GM can be applied to show that $16xy \ge 2x^{5/2}y^{5/2} \implies 8 \ge (xy)^{3/2} \ge xy$. So, a very small number of cases to check. We will find the solution $(2,2)$ which in turns means $(-2,2)$ is a solution to the original equation.

Last case is only $y$ is negative but that's obviously impossible.

Hence $(0,0)$ and $(-2,2)$ are the only possible pairs satisfying the given relation

The following is neither intuitive nor simple, but it does give a different approach to the proof.

If $xy\not=0$, let $p$ be an odd prime and write $x=p^ru$ and $y=p^sv$ with $p\not\mid uv$. From $p^{5r}u^5-p^{5s}v^5=16uvp^{r+s}$, we see we cannot have $r=s\not=0$, so we either have $5r=r+s$ or $5s=r+s$. This means that we can write $x$ and $y$ in the form $x=2^aA^4B$ and $y=2^bAB^4$ with $A$ and $B$ relatively prime odd numbers. But we now have $2^{5a}A^{20}B^5-2^{5b}A^5B^{20}=2^{a+b+4}A^5B^5$, from which we obtain

$$2^{5a}A^{15}-2^{5b}B^{15}=2^{a+b+4}$$

so we must now have $a=b$ (since otherwise the left hand side factors into a power of $2$ times an odd number not equal to $1$), which implies $2^{5a}(A^{15}+B^{15})=2^{2a+4}$, or

$$2^{3a}(A^{15}-B^{15})=2^4$$

The only $15$th powers of odd numbers that differ by a small power of $2$ come from $A=1$ and $B=-1$, so the only solution with $xy\not=0$ is $x=-2$ and $y=2$.