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 $m&lt;x$ 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 $n &lt; y$. Then $p^3 (m^5 + m^2 n^3 p + n^5 p^2) =0$ and now $m^5 + m^2 n^3 p + n^5 p^2=0.$
Your second attempt looks good (the moment you note explicitly that $(m,n)$ is also a solution), and leads to a contradiction in very much the same way as the standard proof that $x^2-2y^2=0$ has no non-zero integer solution (leading to $\sqrt2$ being irrational).
Above equation shown below:

$x^5+px^2y^3+p^2y^5 = 0$  -------$(1)$

Equation $(1)$ has numerical solution given below:

$(x, y, p)=(-6, 3, -8)$