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

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?

1 Answer

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 $$4$$).

• Where did you use quadratic reciprocity in the solution? Commented Nov 1, 2020 at 19:24
• @SARTHAKGUPTA, since $5\equiv1$ mod $4$, quadratic reciprocity says $({5\over M_\ell})=({M_\ell\over 5})$, The OP had already taken that step. Commented Nov 1, 2020 at 19:39