Recent content by RichardCypher

The square of any natural number mod 4 has to be 0 or 1. Therefore, the sum of two such squares mod 4 has to be 0, 1, or 2. However, out of four consecutive natural numbers mod 4, one has to be 3. Contradiction. Is that right? Great hint! Thank-you very much :biggrin:

Hi everybody :smile: I'm currently reading Burton's Elementary Number Theory (almost done!) and in the chapter about Lagrange's Theorem about the sum of four squares, there is a supposedly easy question which I can't solve for some reason :blushing:. I'd really appreciate a hint or two...

I get it! Is it because \phi(5) = 4 and therefore the denominator is equivalent to 0 (mod 5) ? It's only so with p=5!:biggrin: Hurkyl, thank you so much! Your hints are perfect, exactly in the suitable level. They helped me get to it "by myself" but weren't so easy as to not make me think at...

Dear Hurkyl, thank you very much for all your help :biggrin:. There is just a tiny detail I don't quite get. Why wouldn't the exact same proof work for p=5? Thanks again! R.C.

r is a primitive root of p and therefore r^(p-1)=1(modp) so the numerator is 0 and the whole thing is 0 mod p, meaning p|S. Is that right? Is that it?:biggrin: [you're right, of course, it should be r^2 in the numerator]

Thank-you Hurkyl, for your time :smile: I did try to do it, of course, I got \frac{r^4(r^{2(p-1)}-1)}{r^4-1} and couldn't get rid of the fraction :frown:

I remembered a thing that may help: If r is a primitive root of p, then the quadratic residues of p are equivalent to the even powers of r (that is r^2, r^4, \dots, r^{p-1}) and the quadratic nonresidues to the odd powers of r. So, basically, we want to prove that p|r^2 + r^6 + r^{10} +...

Not so. I was wrong. We can multiply the formula by (2n+1)/3 provided that it's an arithmetic progression. So it won't necessarily work in our case. Hmmpf, I don't have any idea how to prove that p divides the sum of squares of its quadratic residues, given that it divides the sum of its...

You are correct, it would follow. However, I am not sure how to prove such a thing. It is easy enough to show that in order to get to the formula of the sum of the squares from the formula of the "regular" sum, one needs only to multiply by a factor of (2n+1)/3, or p/3, in our case. Meaning that...

Morning huba :smile: Thanks for your reply. Hmmpf, I think it can be shown that for any prime, p>3, p divides the sum of its quadratic residues. For example, it's a well-known fact that the quadratic residues of p equals to 1^2, 2^2, \dots, \left(\frac{p-1}{2}\right)^2 \mbox{ (mod } p)...

Morning everybody!:smile: I'm trying to prove a curious little property of quadratic nonresidues: Let p>5 be a prime number. p|S where S is the sum of the squares of its [p's, that is] quadratic nonresidues. However, I didn't make any headway with this problem. I will very much appreciate...

OK, I got it! Bunch of thanks for your help!:biggrin:

thanks for the reply :smile: if r is a primitve root then r^[(p-1)/2] is -1 (mod p). However, (-r)^[(p-1)/2] is the same since (p-1)/2 is even, am I right? Where do we go from here? Maybe I could say that the order of (-r) couldn't be lower than (p-1), because if it was so, ord(-r)|(p-1)/2...

Let r be a primitive root of a prime number p \geq 3. Prove that if p \equiv 1 (mod 4), then -r is also a primitive root of p. I've been told it's quite easy, but I can't see why it's true for the life of me :frown: