Existence of Non-Square Residue

Theorem
Let $m \in \Z$ be an integer such that $m > 2$.

Let $\Z_m$ be the set of integers modulo $m$:
 * $\Z_m = \left\{{\left[\!\left[{0}\right]\!\right]_m, \left[\!\left[{1}\right]\!\right]_m, \ldots, \left[\!\left[{m - 1}\right]\!\right]_m}\right\}$

Then there exists at least one residue in $\Z_m$ which is not the product modulo $m$ of a residue with itself:
 * $\exists p \in \Z_M: \forall x \in \Z_m: x \cdot_m x \ne p$

Proof
We have that $1 \in \Z_m$ and that:
 * $1 \cdot_m 1 = 1$

We have that $m - 1 \in \Z_m$ and that:
 * $\left({m - 1}\right) \cdot_m \left({m - 1}\right) = 1$

Thus unless $m - 1 = 1$, that is, $m = 2$, there exist $2$ residues of $\Z_m$ whose product modulo $m$ with itself equals $1$.

There are $m - 2$ residues which, when multiplied modulo $m$ with themselves have as a result a residue.

Thus there can be at maximum $m - 2$ residues (excluding $1$) which can be the product modulo $m$ of a residue with itself.

But there are $m - 1$ residues (excluding $1$).

So at least $1$ residues is not the product modulo $m$ of a residue with itself.