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I try to prove that $3$ is a quadratic non residue for Mersenne prime.

So, I need to prove that $(\frac{3}{p})=-1$. It happens when $q\equiv5,7\pmod{12}$. But I don't know how to prove that $2^p-1=q\equiv5,7\pmod{12}$...

I also try to check if 3 is always a primitive root for Mersenne primes. For $2^p+1$ is easy but what about $2^p-1$?

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up vote 3 down vote accepted

Let $p=2^q-1$ where $q$ must be prime for $p$ to be Mersenne prime.

If $q=2,$ $p=2^2-1=3$

So, we can start with odd $q>2$

$\implies \frac{p-1}2=\frac{2^q-2}2=2^{q-1}-1$ which is odd if $q-1\ge1$

$$\left(\frac3 p\right)\left(\frac p3 \right)=(-1)^\frac{(3-1)(p-1)}4=-1$$

As $q$ is odd $q=2r+1$(say)

$2^q=2\cdot2^{2r}=2\cdot4^r\equiv2\pmod 3$ as $(4-1)\mid(4^r-1)$

$\implies p=2^q-1\equiv1\pmod 3$

$\implies \left(\frac p3 \right)=\left(\frac 13 \right)=1$

$\implies \left(\frac3 p\right)=-1$

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If $p>2,2^p=0\pmod4,2^p=(-1)^p=-1\pmod 3$,so $2^p=8\pmod{12},2^p-1=7\pmod{12}$

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I guess you mean $2^p=8 (\operatorname{mod} 12)$ not (mod 1)2 – Dominic Michaelis Mar 31 '13 at 10:13

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