Squares in a factorring of Z

Question

I'm stucked at some exercise:

Show that at least one of the three integers {-2,-1,2} is a square in $\mathbb{Z}/(p)$ where p is a prime number.

I have no ideas so far and happy about all kind of help.

Answer 1

Hint: The multiplicative group of the finite field $\mathbb{F}_p$ is cyclic. Let $x$ be a generator. Then $n$ is a square in $\mathbb{F}_p^*$ if and only if it is an even power of $x$. Suppose that $-2$ and $-1$ are not squares. Then we have $x^{2k+1}\equiv -2\bmod p$ and $x^{2l+1}\equiv -1 \bmod p$. But then $$ 2=(-2)(-1)\equiv x^{2(k+l+1)} \bmod p $$ is a square.

More general we have:

Let $F$ be a finite field. If $a, b\in F$ are both non-squares, then $ab$ is a square.

Proof: This duplicate.

Answer 2

Here is a proof that is due to Gauss when $p$ is an odd prime number. By a "residue modulo $p$," we mean an integer $x$ which is not divisible by $p$ which is congruent modulo $p$ to the square of an integer. A "nonresidue" is an integer which is not divisible by $p$ and which is also not a residue.

Since the $p$ elements $$-\frac{p-1}{2}, ... , -2,-1,0,1,2, ... \frac{p-1}{2}$$

constitute representatives for the elements of $\mathbb{Z}/(p)$, it is clear that no more than half of the elements $1, ... , p-1$ are squares modulo $p$. This is because all such squares will be found modulo $p$ among the elements $1^2, 2^2 ,... [\frac{p-1}{2}]^2$.

But actually, exactly half of them will be squares. To see this, we just need to check that if $0 < a < b \leq \frac{p-1}{2}$, then $a^2$ and $b^2$ are not congruent modulo $p$. But if $a^2 \equiv b^2 \pmod{p}$, then $p$ divides $(b-a)(a+b)$, hence $p$ divides $b-a$ or $a+b$. This is impossible, since both of these numbers are positive and less than $p$.

It follows from here that:

1 . The product of a nonresidue and a residue is a nonresidue

2 . The product of two nonresidues is a residue.

and this implies that $-2 = (-1) \cdot 2$ must be a residue if $-1$ and $2$ are not.

Proof: (1) is pretty easy: if $a \equiv x^2$ and $b$ is a nonresidue, we can suppose by way of contradiction that $ab \equiv y^2$ for some integer $y$. Since $a$ is not divisible by $p$, neither is $x$, so there exists an integer $z$ such that $xz \equiv 1$. Then $b \equiv (zy)^2$, contradiction.

For (2), let $a, b$ be nonresidues modulo $p$, and let $x_1, ... , x_{\frac{p-1}{2}}$ be all the residues. Then $ax_1, ... , ax_{\frac{p-1}{2}}$ constitute all $\frac{p-1}{2}$ of the nonresidues by (1). If we suppose by way of contradiction that $ab$ is still a nonresidue, then we would have $ab \equiv ax_i$ for some $i$, which implies $b \equiv x_i$, a contradiction. $\blacksquare$