A system of quadratic diophantine equations Find $(p,x,y)$, where $p$ is prime and $x,y$ natural numbers, such that:
$p+1=2x^2$
$p^2+1=2y^2$
A solution to this problem is $(7,2,5)$ and I tried to show that it is the only one, but I didn't get far. Can anyone help?
First of all, rule out $p\in\mathbb \{2,3\}$. Hence, $2\mid p-1$. Then, observe that your equations yield $$p\cdot \frac{p-1}2=(y-x)(y+x)$$ We could have $p\mid y-x$, but since $p&gt;\frac{p-1}2$, this would imply that $y-x&gt;y+x$. Contradiction. Thus, we have $p\mid x+y\implies p\leqslant x+y\implies 2p\leqslant 2(x+y)$. At the same time, we infer that $p-1\geqslant 2(y-x)$.
Add these inequalities to obtain $4x\geqslant p+1\iff 4x\geqslant 2x^2\iff x\in\mathbb \{0,1,2\}$. It is easy to see that $(x,y,p)\equiv (2,5,7)$ is the only solution.
From $p+1=2x^2, p^2+1=2y^2$ we have that $$p(p-1)=2(y-x)(y+x).$$ Since $p$ is prime it is $y-x=1$ or $y+x=p.$

If $y-x=1$ then $p=3$ and there is no solution.
If $y+x=p$ then $p+1=4x=2x^2\implies x=2.$ And thus $y=5,p=7.$