1. ## An Interesting Discovery

I found something fascinating, the number of incongruent quadradic residues of a prime p>2 divides
(p-1). The number of cubic residues of a prime p divides (p-1). The number of quartic residues of a prime p divides (p-1). This continues indefinitely.
I proved this like this, the set of all nth residues of a prime p, forms a group under multiplication modulo p (I showed an inverse exits by the Pigeonhole Principle)-furthermore this is a cyclic group-and by Lagrange's Theorem the order of a subgroup (finite) divides the order of the group (finite). Thus L(n) divides (p-1) where L(n) is the number of incongruent n'th residues.

2. I'm surprised no one else commented on this. I think that's a nice result. My guess is that its probably well known, but still, its a cool result.

I tried proving that every quadratic residue had an inverse like you said, by some kind of counting argument, but I wasn't getting anywhere. Instead I noticed that since the group of units mod p is an abelian group, the function

f(a) = a^(-1)

is an isomorphism. Then if a is a quad residue, let b be such that b^2 == a mod p. then

f(a) = f(b^2) = f(b)^2 = a^(-1)

So that a^(-1) is congruent to f(b)^2, i.e. b^(-2).

3. I proved it like this: