Find all the integer pairs $(x, y)$ which satisfy the equation $x^5-y^5=16xy$ I just came across the following question:

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

I solved it as follows:
$x=y=0$ obvious solution. If $xy\neq0$, let $d=gcd(x, y)$ and we write $x=da$, $y=db$, $a, b\in \Bbb{Z}$ with $(a, b)=1$. Then, the given equation is:
$$d^3a^5-d^3b^5=16ab$$
So, by the above equation, $a$ divides $d^3b^5$ and hence $a$ divides $d^3$. Similarly $b$ divides $d^3$. Since $(a, b)=1$  we have that $ab$ divides $d^3$, so $d^3=abr$ with $r\in \Bbb{Z}$. Then the above equation becomes $abra^5-abrb^5=16ab$, so $r(a^5-b^5)=16$.
Hence, the difference $a^5-b^5$ must divide $16$. If $|(a^5-b^5)|\le2$ we have that $(x, y)=(-2, 2)$ is a solution. Otherwise $$|a^5-b^5|=|(x+1)^5-b^5|\ge |(x+1)^5-x^5|=|5x^4+10x^3+10x^2+5x+1|\ge31$$ which is impossible.
So only solutions are $(x, y)=(0, 0)$ or $(-2, 2)$.
I believe that this solution is not at all intuitive nor simple. Could you please post a more intuitive and simple solution where you are explaining your intuition on every step?
 A: 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
A: 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 $y=2$.
