Square Modulo 5

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Theorem

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

Then one of the following holds:

\(\ds x^2\) \(\equiv\) \(\ds 0 \pmod 5\)
\(\ds x^2\) \(\equiv\) \(\ds 1 \pmod 5\)
\(\ds x^2\) \(\equiv\) \(\ds 4 \pmod 5\)


Corollary

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


Proof

Let $x$ be an integer.

Using Congruence of Powers throughout, we make use of $x \equiv y \pmod 5 \implies x^2 \equiv y^2 \pmod 5$.

There are five cases to consider:

$x \equiv 0 \pmod 5$: we have $x^2 \equiv 0^2 \pmod 5 \equiv 0 \pmod 5$.
$x \equiv 1 \pmod 5$: we have $x^2 \equiv 1^2 \pmod 5 \equiv 1 \pmod 5$.
$x \equiv 2 \pmod 5$: we have $x^2 \equiv 2^2 \pmod 5 \equiv 4 \pmod 5$.
$x \equiv 3 \pmod 5$: we have $x^2 \equiv 3^2 \pmod 5 \equiv 4 \pmod 5$.
$x \equiv 4 \pmod 5$: we have $x^2 \equiv 4^2 \pmod 5 \equiv 1 \pmod 5$.

$\blacksquare$


Sources