Show that if p is an odd prime and a∈Z such that p doesn't divide a then x2≡a(modp) has no solutions or exactly 2 incongruent solutions.
The only theorem that I thought could help was:
Let a,b and m be integers with m>0 and gcd(a,m)=d. If d doesn't divides b then ax≡b(modp) has no solutions. If d divides b then ax≡b(modp) has exactly d incongruent solutions modulo m.
But I feel that this would be an invalid theorem for this proof since ax and x2 are of different degrees.
Thoughts? Other methods to approach this?
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