Square Modulo 5

Theorem
Let $x \in \Z$ be an integer.

Then one of the following holds:

Corollary
When written in conventional base 10 notation, no square number ever ends in one of $2, 3, 7, 8$.

Proof
Let $x$ be an integer.

Using Congruence of Powers throughout, we make use of $x \equiv y \ \left({\bmod 5}\right) \implies x^2 \equiv y^2 \ \left({\bmod 5}\right)$.

There are five cases to consider:


 * $x \equiv 0 \ \left({\bmod 5}\right)$: we have $x^2 \equiv 0^2 \ \left({\bmod 5}\right) \equiv 0 \ \left({\bmod 5}\right)$.


 * $x \equiv 1 \ \left({\bmod 5}\right)$: we have $x^2 \equiv 1^2 \ \left({\bmod 5}\right) \equiv 1 \ \left({\bmod 5}\right)$.


 * $x \equiv 2 \ \left({\bmod 5}\right)$: we have $x^2 \equiv 2^2 \ \left({\bmod 5}\right) \equiv 4 \ \left({\bmod 5}\right)$.


 * $x \equiv 3 \ \left({\bmod 5}\right)$: we have $x^2 \equiv 3^2 \ \left({\bmod 5}\right) \equiv 4 \ \left({\bmod 5}\right)$.


 * $x \equiv 4 \ \left({\bmod 5}\right)$: we have $x^2 \equiv 4^2 \ \left({\bmod 5}\right) \equiv 1 \ \left({\bmod 5}\right)$.

Proof of Corollary
The absence of $2$ and $3$ from the digit that can end a square follows directly from the above.

As $7 \equiv 2 \ \left({\bmod 5}\right)$ and $8 \equiv 3 \ \left({\bmod 5}\right)$, the result for $7$ and $8$ follows directly.