Working with finite fields of form Z_p

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The discussion centers on proving that not every element in the finite field Z_p (where p is an odd prime) is a square of another element. Initial attempts at proof were flawed due to misinterpretation of congruences. Participants emphasized the importance of understanding the structure of Z_p, particularly that the group of units is cyclic and that squaring is not a one-to-one function. Examples from Z_3 and Z_5 illustrate that certain elements, like 2 and 3, are not squares. The conclusion drawn is that counting arguments show there aren't enough distinct square roots to cover all elements in Z_p.
Syrus
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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.

Homework Equations


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?
 
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In a finite field, the group of units is cyclic. This gives you a quick way of proving the result.
 
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.
 
Syrus said:
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?
 
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.
 
Last edited:
Syrus said:
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.

Ok, I think you got what Z_p means. Now use counting logic. If every element of Z_p were a square then for every element z of Z_p there is an element k such that k^2=z. If 1^2=1 and (-1)^2=(p-1)^2=1 do you have enough elements of Z_p to give each element a square root?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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