I'm reading "SQUARE FIBONACCI NUMBERS, ETC." by JOHN H.E. COHN. (http://math.la.asu.edu/~checkman/SquareFibonacci.html). There is a portion of it in theorem 1 that I do not understand:

$L_n \text { is the nth term of the Lucas sequence}$.

k is any even integer that is not divisible by 3.

It is proven that $L_n \equiv -1 (\text{mod }L_k)$ and $L_k \equiv 3 (\text{mod } 4)$

The author concludes from the above that -1 is a non-residue of $L_k$.

From reading up on quadratic residues, all the methods of determining if x is a non residue of p seem to depend on p being an odd prime. However in the above case, $L_k$ is not always an odd prime. This is probably a trivial question, sorry in advance, I have read two textbooks on this and can find no way of proving that -1 is a non-residue of $L_k$

Thanks for any help

I can't answer my own question cause i'm new, here is the answer suggested by Gerry Myerson below.

First off:

If p is a prime number and p≡3(mod 4), then -1 is a non residue modulo of p (law of quadratic reciprocity, first supplement)

Then:

Since n≡3(mod 4), n has a prime factor q for which q ≡3(mod 4). Since q is a prime number and q≡3(mod 4), -1 is a non residue modulo of q. Since n is a multiple of q, -1 is also a non residue modulo of n.

• If $n\equiv3\pmod4$ then $n$ has a prime factor that is also 3 (mod 4). And $-1$ is a nonresidue modulo any such prime, hence, modulo $n$. – Gerry Myerson Nov 1 '13 at 10:06
• That's because it's a comment, not an answer. But if you understand it, you can write it up in your own words and post it as an answer, and then later you can accept it. – Gerry Myerson Nov 1 '13 at 10:13
• Okay sorry, i'm new to this. – JeremyT Nov 1 '13 at 10:14