Solutions for $p$ where $2 p^2 - 1 = q^2$ Consider this equation:
$$2 p^2 - 1 = q^2$$
where $p$ and $q$ are prime.
After vigorous checking, I couldn't find any solutions $p>29.$
Is it so that $p=5,\;p = 29$ are the only solutions?
Edit:
Found solutions $p>29$ with the help this.
The solution for $p,q$ can be expressed as:
$$\begin{align}p &=a_n= {1\over4} \left(2 (3-2 \sqrt{2})^n+\sqrt{2} (3-2 \sqrt{2})^n+2 (3+2 \sqrt{2})^n-\sqrt{2} (3+2 \sqrt{2})^n\right)&\\
q &=b_n=- {1\over2} \left((3-2 \sqrt{2})^n+\sqrt{2} (3-2 \sqrt{2})^n+ (3+2 \sqrt{2})^n-\sqrt{2} (3+2 \sqrt{2})^n\right)&\end{align} $$
thus:
$$\begin{align}
2 p^2 - 1 = q^2\rightarrow\text{True}&\implies\\
p = a_n,\;q=b_n,1<n\\
\end{align}$$
So when both $a_n$ and $b_n$ are both primes, then $2 a_n^2 - 1 = b_n^2$.
Solutions for $n$ I've found so far are:
$$\begin{align}
n=2&&a_n = 5&&b_n = 7\\
n=3&&a_n = 29&&b_n = 49\\
n=15&&a_n = 44560482149&&b_n = 63018038201\\
n=30&&a_n = 19175002942688032928599&&b_n = 19175002942688032928599\\
\end{align}$$

There are at least 2 more solutions, you just haven't searched far enough
OEIS A163742 gives solutions $(63018038201, 44560482149)$ and $(19175002942688032928599, 13558774610046711780701)$. It is conjectured that these are the only 4 prime pairs.
