Suppose for some then since thus: if then , but since thus .

By Fermat's Little Theorem: and hence i.e. .

So there can be no solutions when .

Suppose , then

Now, by Wilson's Theorem , and since we have , thus and a solution exists.

The fact that there are exactly 2 solutions to follows easily from the fact that if you have then either or

- I suppose you have not gone over quadratics residues, just as a comment this will get much more generalised when you get there.