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Suppose the $p$ is an odd prime not dividing $ab$. Prove that $z^2 \equiv ab \pmod p$ is solvable if and only if both or neither of $x^2\equiv a\pmod{p}$ and $x^2 \equiv b \pmod p$ are solvable.

orangeskid
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atl
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    Well certainly if both $a$ and $b$ are perfect squares $\pmod{p}$, then so too will be their product. Write out precisely what that means and apply commutativity to make that direction come out. – Kaj Hansen Apr 14 '15 at 17:28
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    Let $\mathbb{Z}_p^* $ be the set of nonzero elements of $\mathbb{Z}/p\mathbb{Z}$. Did you know that $\mathbb{Z}_p^* $ is a cyclic group? Let $g$ be the generator of that group. What happens if you raise $g$ to an even power or odd power? – wlad Apr 14 '15 at 17:33

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In other contexts, one might phrase this question as trying to prove that the product of two quadratic nonresidues is a quadratic residue. The "standard" way typically presented in elementary number theory texts (and suggested by user in the comments) is to first show that the nonzero elements mod $p$ have a generator (a primitive root) $g$. In this case, the squares are precisely those represented by even powers of $g$. Since the product of two odd powers of $g$ is an even power of $g$, that would prove your claim.

But we can go from first principles and use only basic algebra. Denote the nonzero elements mod $p$ by $(\mathbb{Z}/p\mathbb{Z})^\times$, and denote the squares by $S$. Then the key idea is that $S$ is a subgroup of $(\mathbb{Z}/p\mathbb{Z})^\times$.

We can check this directly. If $x \equiv a^2$ and $y \equiv b^2$, then $xy \equiv (ab)^2$. Similarly, $x^{-1} \equiv (a^{-1})^2$, and so $S$ is a subgroup.

We ask the fundamental question: how large is $S$? Since $1^2 = (-1)^2 = 1$, we can't hit every element. Consider the group homomorphism $\sigma: (\mathbb{Z}/p\mathbb{Z})^\times \longrightarrow (\mathbb{Z}/p\mathbb{Z})^\times$ given by $\sigma:x \mapsto x^2$. That is, consider the squaring map. More specifically, consider the kernel of $\sigma$.

We want to know when $\sigma (x) = 1$, or rather $x^2 \equiv 1 \pmod p$. This is equivalent to $p \mid x^2 - 1 = (x - 1)(x + 1)$, or rather $x \equiv \pm 1 \pmod p$. (We have used that $p$ is prime). So the size of the kernel is $2$.

Since very clearly $\sigma$ surjects onto $S$, we see that $S$ is an index $2$ subgroup of $(\mathbb{Z}/p\mathbb{Z})^\times$. An index $2$ subgroup is always normal, and so we can make sense of the quotient group $(\mathbb{Z}/p\mathbb{Z})^\times / S$.

This quotient group has $2$ elements. There is only one $2$ element group, and it's isomorphic to $\{1, -1\}$ under multiplication. Being $1$ in this quotient group corresponds to an element of $S$. Being $-1$ corresponds to being a non-square.

It is now clear that the product of two nonsquares, in the quotient group, acts like $-1 \cdot -1 = 1$, and is therefore a square.

More generally, the product of two squares is a square, the product of two nonsquares is a square, and the product of a square and a nonsquare is a nonsquare. It behaves (quite literally) like $-1$ and $1$ under multiplication. $\diamondsuit$

davidlowryduda
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