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Show that for $p\neq2$, not every element in $\mathbb{Z}/p\mathbb{Z}$ is a square of an element in $\mathbb{Z}/p\mathbb{Z}$. (Hint: $1^2=(p-1)^2=1$. Deduce the desired conclusion by counting).

So far I have that $1=p^2-2p-1\Rightarrow p^2-2p=0\Rightarrow p^2=2p$, but I don't know where to go from here. I also don't fully understand what it means to deduce it by counting.

Jendrik Stelzner
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Matt G
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  • Really Relevant: http://math.stackexchange.com/questions/1631849/prove-that-there-are-as-many-quadratic-residues-mod-p-as-there-are-quadratic-non/1632350#1632350 – Maffred Feb 01 '16 at 02:16

3 Answers3

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Because $\mathbb{Z}/p\mathbb{Z}$ is finite the map $f \colon \mathbb{Z}/p\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$, $x \mapsto x^2$ is surjective (i.e. every element is a square) if and only if it is injective. But for $p \neq 2$ we have $-1 \neq 1$ with $f(-1) = 1 = f(1)$.

Jendrik Stelzner
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You're going off in the wrong direction. Since $1^2=(p-1)^2$, the function $f(x)=x^2$ is not a subjection. Therefore its image is a strict subset of $Z/pZ$, since it's a finite set.

Put another way, there are $p$ numbers, but two of them have the same square, so we can go count the squares and find there are less than $p$.

Extending this argument a bit tells you that in fact half of the numbers are squares.

Stella Biderman
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It means you should count how many squares there are, they are certainly less than $p$ since $1^2=-1^2$ (and $1\neq -1$ when $p\neq 2$)

Asinomás
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