# Working with finite fields of form Z_p

Syrus

## Homework Statement

Let p be an odd prime. Then Char(Z_p) is nonzero.

Prove: Not every element of Z_p is the square of some element in Z_p.

## The Attempt at a Solution

I first did this, but i was informed by a peer that it was incorrect because I was treating the congruency as an equality:

Suppose not. Then every element of Z_p is the square of some element in Z_p. Take 1. Since in mod p: 1 = (p-1)2 = 12, it follows that 0 = p2 - 2p, and hence p = 2, a contradiction.

What can i do to find a correct way of proving this?

Gold Member
In a finite field, the group of units is cyclic. This gives you a quick way of proving the result.

Homework Helper
It's probably even easier if you simply observe that if i^2=k then (-i)^2=k. But since your first proof is so bogus, you probably really don't understand what Z_p is. Try doing a warmup and figuring out what elements of Z_3, Z_5 and Z_7 are squares and which aren't.

SteveL27
Suppose not. Then every element of Z_p is the square of some element in Z_p. Take 1. Since in mod p: 1 = (p-1)2 = 12, it follows that 0 = p2 - 2p, and hence p = 2, a contradiction.

Are you allowed to divide by p in Zp?

For example suppose we're in Z5. Now

5 * 2 = 10 = 0 (mod 5)

and 5 * 3 = 15 = 0 (mod 5).

So since 5*2 = 5*3 we conclude that 2 = 3.

What's wrong with that proof?

Syrus
Yes, I suppose that's silly... dividing by p (which is congruent to zero). I have done what dick suggested and discovered that for Z_3, 2 is not a square. For Z_5, neither 2 nor three are squares. I am trying to work off this for now but can't seem to generalize these cases.

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