<s>[INST] Solving $x^5+px^2y^3+p^2y^5 = 0$ where $p$ is prime and $x,y \in \mathbb{Z}$ How do you solve $x^5+px^2y^3+p^2y^5 = 0$ where $p$ is prime and $x,y \in \mathbb{Z}$?

Working
 in modulo $p$  we have $x^5 = 0 \pmod{p}$ and $x = 0$ which is the only solution in modulo $p$ since $a^{p} \equiv a \mod{p}$ for any prime $p$ and integer $a$. Taking this back we get $p^2y^5 = 0$ which implies $y=0$. Hence the only solution is $(x,y) = (0,0)$.  

Edit/attempt 2: 
Since $x = 0 \mod{p}$ we can let $x = pm$ where $m<x$ then we get  $p^2 (m^5 p^3 + m^2 p y^3 + y^5) = 0 \implies m^5 p^3 + m^2 p y^3 + y^5$ since $p$ can't be zero. But now $y^5 = 0 \mod{p}$ ; suppose $y = pn$ where $n < y$. Then $p^3 (m^5 + m^2 n^3 p + n^5 p^2) =0$ and now $m^5 + m^2 n^3 p + n^5 p^2=0.$
 [/INST] I would like to emphasize the fact that the equation is a homogeneous polynomial in x,y, all "total" exponents are 5.  What that means is that, if we began with $\gcd(x,y) = g > 1,$ we are allowed to doivide both $x,y$ by the same $g,$ arriving at a coprime pair still giving a solution. 
In contrast: if we began with $p s^3 = p^2 t^2,$ variables $s,t,$ we get equality with $s=p, t=p$ or with $s=p^3, t=p^4,$ but no solution with $s,t$ coprime. Generally $s = p^{1+2k}, t = p^{1+3k}$ as well as $s = a^2 p^{1+2k}, t = a^3 p^{1+3k}$
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