<s>[INST] Primes of the form $p=2a^2-1$ satisfying $p^2=2b^2-1$. Are there any prime numbers $p$ for which there exist integers $a$ and $b$ such that
$$p=2a^2-1\qquad\text{ and }\qquad p^2=2b^2-1,$$
other than $p=7$?
The fact that $p^2=2b^2-1$ implies that
$$(p+b\sqrt{2})(p-\sqrt{2}b)=-1,$$
and hence that $p+b\sqrt{2}$ is a unit in $\Bbb{Z}[\sqrt{2}]^{\times}=\langle-1,1+\sqrt{2}\rangle$, from which it quickly follows that 
$$p+b\sqrt{2}=(1+\sqrt{2})^{2n+1},$$
for some integer $n\geq0$. This line of thought shows that $p$ must be of the form
\begin{eqnarray*}
p&=&\frac{(1+\sqrt{2})^{2n+1}+(1-\sqrt{2})^{2n+1}}{2},\\
p&=&\begin{pmatrix}1\\0\end{pmatrix}
\begin{pmatrix}0&1\\-1&6\end{pmatrix}^n\begin{pmatrix}1\\7\end{pmatrix},\\
p&=&a_n\quad\text{ where }\quad a_0=1, a_1=7\quad\text{ and }\quad a_{n+1}=6a_n-a_{n-1},
\end{eqnarray*}
for some integer $n\geq0$.
Checking up to $n=150$ yields no solutions, but unfortunately I'm not computer-savvy enough to check quickly. I am also at a loss how to prove that there are no other such primes. Any ideas are welcome.
 [/INST] Clearly $p\ne 2$ and $p>b>a$. Write $$ p(p-1)= 2(b-a)(a+b)$$
Case 1: $$p\mid a+b\implies a+b = kp \implies k<2$$
So $a+b=p$ and $p-1 = 2(b-a)$. From here we get $p=4a-1$ so $$4a-1 = 2a^2-1\implies a=2$$
so $p= 7$.  
Case 2: $$p\mid b-a\implies p \leq b-a <p $$
Which is impossible.
</s>