<s>[INST] $p+7=2\cdot x^2$ and $p^2+7=2\cdot y^2$. Let $p$ a prime number such that there exists $x,y$ positive integers such that $p+7=2\cdot x^2$ and $p^2+7=2\cdot y^2$.
Find $p$.
I found that $p=11$ . I noticed that $p|(x+y)$. I tried with quadratic residues but I am stuck.
 [/INST] Subtracting the second problem equation minus the first one gives
$$p^2 - p = 2y^2 - 2x^2 \; \; \to \; \; p(p - 1) = 2(y - x)(y + x) \tag{1}\label{eq1A}$$
Note $p$ being an odd prime means $p \mid y - x$ or $p \mid y + x$. The first case gives $y - x \ge p$ and $y + x \gt p$, so $p^2 - p = 2(y - x)(y + x) \gt 2p^2$, which is not possible. Thus, as you already deduced, this means
$$p \mid x + y \tag{2}\label{eq2A}$$
If $y \ge p \; \to \; 2y^2 \ge 2p^2$, then the second equation gives $p^2 + 7 \ge 2p^2 \; \to \; p^2 \le 7$, but $p \ge 3$, which means this is not possible. Thus, $y \lt p$, and since $p + 7 \lt p^2 + 7$, then $x \lt y \; \to \; x \lt p$, so
$$x + y \lt 2p \; \; \to \; \; x + y = p \; \; \to \; \; y - x = p - 2x \tag{3}\label{eq3A}$$
Substitute this into the RHS of \eqref{eq1A} to get
$$\begin{equation}\begin{aligned}
p(p - 1) & = 2(p - 2x)p \\
p - 1 & = 2p - 4x \\
p & = 4x - 1
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Using this in the problem's first equation gives
$$\begin{equation}\begin{aligned}
4x + 6 & = 2x^2 \\
x^2 - 2x - 3 & = 0 \\
(x - 3)(x + 1) & = 0
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
With $x$ being a positive integer, then $x = 3$, so using \eqref{eq4A} gives your one result of $p = 11$.
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