Find all primes $p$ such that $(2^{p-1}-1)/p$ is a perfect square. I tried brute-force method and tried to find some pattern. I got $p=3,7$ as solutions. Apart from these I have tried for many other primes but couldn't find any other such prime.
Are these the only primes that satisfy the condition?
If yes, how to prove it theoretically and if not, how to find others?
Thanks in advance!
Best Answer
Hint: Let $p=2k+1$ where $k \in \mathbb{N},$ then $2^{2k}-1=(2^k-1)(2^k+1)=pm^2.$ We know that $\gcd(2^k-1,2^k+1)=1$ since they are consecutive odd integers, so the equation breaks into $2^k-1=px^2, 2^k+1=y^2$ or $2^k-1=x^2, 2^k+1=py^2.$
Easy investigation shows that the only solutions are $p=3,7.$ I will leave it to you to fill the gaps. You may also interested in this.