Find $p$ if $2p-1$ is a perfect square

Question

I need to find prime numbers $p$ such that $2p-1$ is a perfect square. I tried hard, but could not get a proper solution for this. I could guess $p=13$ works, but I need a proper rigorous solution for this. I cant think of one.

Answer 1

All primes excluding $2$ and $3$ are of the form $6k\pm 1$. We have that:

$$2(6k+1)-1=12k+1$$ This leads to $$12k=m^2-1$$ $$12k=(m+1)(m-1)$$ $$\to m+1=12a, m-1=\frac ka$$ Or vice versa, subject to $a,k \in \Bbb Z, a|k$

Now use: $$2(6k-1)-1=12k-3$$ $$\to 12k-4=m^2-1$$ $$\to 4(3k-1)=(m+1)(m-1)$$ Go from there.

Answer 2

OEIS A027862 lists the primes $p$ of the form $p=n^2+(n+1)^2$. It remarks that these are the ones where $2p-1$ is a square. We can show that if $p$ is of this form $2p-1$ is a square because $$2p-1=2(n^2+(n+1)^2)-1=4n^2+4n+1=(2n+1)^2$$