Square Modulo 5
Jump to navigation
Jump to search
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
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 14$: Worked Example $1$: Congruence modulo $m$ ($m \in \N$)