Condition for $8p+1$ divides $2^p-1$?

Question

Here is what I observed :

Let $8p+1 = (2a-1)^2+64(2b-1)^2$ with $a$ and $b$ be a positive integers, $p$ and $8p+1$ both prime numbers.

Then $8p+1$ divides $2^p-1$ only if you can write $8p+1$ as $(2a-1)^2+64(2b-1)^2$.

For example :

• $89 = (2 \cdot 3 - 1)^2+64(2 \cdot 1 - 1)^2$ and $89 = 11 \cdot 8+1$ so $89$ divides $2^{11}-1$
• $233 = (2 \cdot 7 - 1)^2+64(2 \cdot 1 - 1)^2$ and $233 = 29 \cdot 8+1$ so $233$ divides $2^{29}-1$
• $3449 = (2 \cdot 22 - 1)^2+64(2 \cdot 3 - 1)^2$ and $3449 = 431 \cdot 8+1$ so $3449$ divides $2^{431}-1$
• $137 = (8 \cdot 17 + 1)$ but you can't write $137$ as $(2a-1)^2+64(2b-1)^2$ so $137$ does not divide $2^{17}-1$

For the moment, I didn't find a counterexample with this condition.

I need help for proving it but I don't know how to start.

I thought about Mersenne numbers and Sophie Germain primes that say if $p \equiv 3 \pmod{4}$ and $2p+1$ is prime then $2p+1$ divides $2^p-1$ but for $8p+1$ it doesn't work.

If you found a counterexample please tell me.

Answer 1

I don't know if it can be useful to find a proof but if $p$ and $1+8p$ are prime numbers and $2^p-1 \equiv 0 \bmod {(1+8p)}$ then

$2^{p+1} \equiv 2 \bmod {(1+8p)}$

$(2^{\frac{p+1}{4}})^4 \equiv 2 \bmod {(1+8p)}$ if $p \equiv 3\bmod4$

$(2^{\frac{p+3}{4}})^4 \equiv 8 \bmod {(1+8p)}$ if $p \equiv 1\bmod4$

in the first case, here you will find information on why $1+8p=x^2+64 y^2$

Answer 2

for an odd prime $p$, if a prime $q$ divides $2^p − 1$, then $q \equiv 1$ (mod $2p$). also $q$ $\equiv \pm 1$ (mod $8$).

in the last example of $M_{17}$ which is a Mersenne Prime, the only valid option for $q$ is the number itself which is $(8*16384) - 1$. for all of these $M_{13}$, $N_{19}$ naturally $q \equiv - 1$ (mod $8$). I don't know if that's a useful place to start from, as in - only for non prime $2^p - 1$, $q \equiv \pm 1$ (mod $8$).

Edit: saw this now, that makes sense. I was thinking of it only working for mersenne numbers that aren't primes, but that only eliminates a subset of $M_p$ (it works for prime $M_n$ too)