Thread: 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.

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).

I proved it like this:
Consider the elements which are quadradic residues mod p
S={a1, a2, ..., an}
Take any "a" which is element of S, and consider the multiples of "a" in S. Thus, consider {a*a1, a*a2,...a*an}
Now non of these are congruent to each other for:
a*ak=a*aj(modp) would imply that ak=aj(modp) which is impossible thus, they cannot. Also realize that * is closed. Thus by the Pigeonhole Principle it is a repetion of set S. Thus there must exist an "ak" such as a*ak=1 because 1 is element of S!!!