Find all the integer solutions $(p, q ∈ \mathbb Z)$ of the equation:


I'm here: $$(4pq+2q^2+p^2)(q-1)=0$$

but i only got $q=1$ and with that $p$ only has solutions $\in\mathbb R$

So, the only solutions i see are $p$ = $0$ , $q$ = $0$

Am i wrong?

Also if i'm right, how can i prove that?


You have already dealt with the case $q-1=0$, so let's deal with the factor $4pq+2q^2+p^2=0$. By the quadratic formula we have $p=(-2\pm\sqrt{2})q$. If $q\in\mathbb{Z}$, the only way that we also have $p\in\mathbb{Z}$ is if $q=0$. This in turn implies $p=0$.

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