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# I have a number theory question.?

Let p be a prime with the property that p=/=1 (mod 5). Prove that the Diophantine equation y^2=x^5 + 1 has exactly p solutions modulo p.

I think I understand how to prove that each diophantine equation has a solution mod p. But I'm pretty lost on this one.

Thanks!

### 1 Answer

- kbLv 77 years agoFavorite Answer
Consider the map f: Zp* ---> Zp* defined by f(x) = x^5 (mod p).

**This is easily checked to be a group homomorphism. (Of course 0 would also map to 0 if needed.)

I claim that this map is 1-1.

ker f = {x in Z/pZ: f(x) = x^5 = 1 (mod 5)}

.......= {1}, since there are d = gcd(5, p-1) = 1 solutions (and p ≠ 1 (mod 5)).

Since f is 1-1 and f is a map from a finite set to itself, f is also onto.

==> The map g: Zp --> Zp (not necessarily a group homomorphism) defined by g(x) = x^5 + 1 is also a bijection, being the composition of f and the bijection sending x to x+1.

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Working mod p:

If x = -1 (mod p), then we have only one solution y = 0 (mod p).

Otherwise, y is nonzero, and there are (p-1)/2 (nonzero) quadratic residues.

For each such y, there are two such elements whose (say n and its opposite) square is y.

Hence, there are 1 + 2 * (p-1)/2 = p solutions modulo p.

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I hope this helps!