Hi please help me on these problems
1. For p an odd prime and a any integer which is not congruent to 0 mod p, prove that the congruence x^2 = -a^2 (mod p) has solutions if and only if p=1 (mod4)
2. For integers a and b with (a,b) =1, prove that if p is any odd prime which divides a^2+b^2 then p=1 (mod4)
Thank you very much in advance!
For number 2, is it just that we have to show neither of a or b are 0 mod p, after which we can just apply question 1?
Since p|a^2+b^b, either p|a and p|b or p divides neither. The former would contradict (a,b)=1. Right?
Thanks
Edit: also, is there a way to prove question 1 without using the legendre symbol?
Does my suggestion work though?
1. For p an odd prime and a any integer which is not congruent to 0 mod p, prove that the congruence x^2 = -a^2 (mod p) has solutions if and only if p=1 (mod4)
2. For integers a and b with (a,b) =1, prove that if p is any odd prime which divides a^2+b^2 then p=1 (mod4)
So to prove problem 2 (that given a^b=-b^2 (mod p) for (a,b)=1, prove p = 1(mod 4)), we can use the "only if" part of problem 1 as long as we can prove that b is not congruent to 0 mod p. And I believe this can be done by saying p|a^2+b^2 -> its impossible for p to divide only one of a or b, and if p could divide both then the GCD of a and b would be at least p (contradicting the given). Thus neither a nor b are congruent to 0 mod p and we can apply "For p an odd prime and a any integer which is not congruent to 0 mod p, the congruence x^2 = -a^2 (mod p) has solutions if and only if p=1 (mod4)".
Does this work?
Also, these are problems for an elementary number theory course and I'm afraid I haven't gotten far enough to understand your new explanation.
Well we know p can't divide b and not a because if for 3 integers x,y, and z, if x|y+z and x|y, then x|z. In this case, p|a^2 + b^2 and if p|b (which means p|b^2), then p|a^2. Since p is a prime, it follows that p|a.
And as I said earlier, if p|a and p|b, then (a,b)=>p, a contradiction. So p doesn't divide a or b.
I got the gist of things now. Thank you very much for your help!