Congruence Proof

Deano10

Show that x^2 + y^2 + 1 ≡ 0 (mod p) is soluble for any odd prime p. Then show that x^2 + y^2 + 1 ≡ 0 (mod m) is soluble for any squarefree odd m.


In the early parts of the exercise, I have so far shown that x^2 ≡ y^2 (mod p) if and only if x ≡ +/- y (mod p).

I have also shown that there are precisely (p+1)/2 integers u in {0,1,...,p-1} such that u ≡ x^2 (mod p) for some x.

Though I am unsure as to how to further proceed. Any help you can offer in how to proceed would be very much appreciated!

Many thanks!