Use the quadratic reciprocity to show that 5 is a square modulo $M_l$ if and only if $l$ ≡ 1 (mod 4).

Question

Let $l$ be an odd prime such that $M_l = 2^l − 1$ is a Mersenne prime. Use the quadratic reciprocity to show that 5 is a square modulo $M_l$ if and only if $l$ ≡ 1 (mod 4).

Here's what I know: Using quadratic reciprocity, we have that $(\frac{5}{M_l}) = (\frac{M_l}{5})$ since 5 ≡ 1 (mod4).

Then by Euler's criterion, we have: $M_l^\frac{5-1}{2}= M_l^2 ≡ (\frac{M_l}{5})$ (mod5).

I've examined $(2^l -1)^2$ with $l$ equal to the first ten odd primes, and have seen that for that set, $(2^l -1)^2 ≡ 1$ (mod5) (which is what we need for 5 to be a square $M_l$) only for primes p ≡ 1 (mod4). I know that this isn't a reasonable proof, as we can't assume that something which holds for a small set of numbers will hold for an infinite set of numbers.

Edit: I have the following theorem in mind: For an element $g$ of order $n$, $g^m = 1$ iff $n|m$. I'm only thinking of this theorem because I've used it to prove that numbers are squares in the past, but I don't see how I could apply it here. Is there something I'm forgetting about the order of Mersenne primes mod p that would help me apply this theorem?

Answer 1

If $\ell\equiv3$ mod $4$, then $2^\ell\equiv3$ mod $5$, in which case $M_\ell=2^\ell-1\equiv2$ mod $5$, which is not a square mod $5$, while if $\ell\equiv1$ mod $4$, then $2^\ell\equiv2$ mod $5$, in which case $M_\ell=2^\ell-1\equiv1$ mod $5$, which is a square. All this stems from the fact that $2^4\equiv1$ mod $5$ (and the fact that an odd number $\ell$ is either $1$ or $3$ mod $4$).