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I've kind of fallen behind in my Algebra class and I really haven't read much about the theory. I'm wondering, how would you go on about proving that if an odd prime, p, does not divide a nor b, but divides the sum of their squares - a^2 + b^2 -, then p = 1 mod 4.

Up to know, I've been considering the special case in which gcd(a,b) = 1. Of course solving for this case solves for the entire problem; also it leaves something to work with since a^2 + b^2 = a + b mod 3 and a^2 + b^2 = 1 mod 4 or a^2 + b^2 = 2 mod 4. From that point all I tried seemingly led to a dead end. Maybe there is a property or something of the kind that I am not aware of and without which the problem becomes very cumbersome?

Up to know, I've been considering the special case in which gcd(a,b) = 1. Of course solving for this case solves for the entire problem; also it leaves something to work with since a^2 + b^2 = a + b mod 3 and a^2 + b^2 = 1 mod 4 or a^2 + b^2 = 2 mod 4. From that point all I tried seemingly led to a dead end. Maybe there is a property or something of the kind that I am not aware of and without which the problem becomes very cumbersome?

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