Find all primes $p$ such that $\dfrac{(2^{p-1}-1)}{p}$ 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!!


Hint: Let $p=2k+1$ where $k \in \mathbb{N},$ then $2^{2k}-1=(2^k-1)(2^k+1)=pm^2.$ We have that $\gcd(2^k-1,2^k+1)=1$ since they are consecutive odd numbers, 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.

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  • $\begingroup$ Very good explanation.. +1. $\endgroup$ – IDOK Jun 22 '12 at 16:35
  • $\begingroup$ @ehsanmo: thanks.your hint really helped. $\endgroup$ – Aang Jun 22 '12 at 18:32
  • $\begingroup$ Glad to hear that :) $\endgroup$ – Ehsan M. Kermani Jun 22 '12 at 18:50
  • $\begingroup$ how to solve this equations? $\endgroup$ – agustin Mar 8 '13 at 7:22
  • 2
    $\begingroup$ @agustin: Take case $1$ when $2^k-1=px^2,2^k+1=y^2$, $\\$ then $y^2-1=2^k\implies(y+1)(y-1)=2^k\implies y+1=2^m,y-1=2^n$ form some $m,n\in \Bbb N\cup\{0\} \\$ which gives $2^m-2^n=2$. Now, the only possible value of $m,n$ is $2,1$ respectively. Therefore, $y=3\implies k=3$. $\\$ Putting $k=3$ in $2^k-1=px^2$ gives $px^2=7$. Since $7$ is a prime $\implies p=7$. Similarly, following case $2$ gives $p=3$ $\endgroup$ – Aang Mar 10 '13 at 6:51

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