Odd Square Modulo 8

Theorem

Let $x \in \Z$ be an odd square.

Then $x \equiv 1 \pmod 8$.

Proof

Let $x \in \Z$ be an odd square.

Then $x = n^2$ where $n$ is also odd.

Thus $n$ can be expressed as $2 k + 1$ for some $k \in \Z$.

Hence:

\(\ds x\)

\(=\)

\(\ds n^2\)

\(\ds \)

\(=\)

\(\ds \paren {2 k + 1}^2\)

\(\ds \)

\(=\)

\(\ds 4 k^2 + 4 k + 1\)

\(\ds \)

\(=\)

\(\ds 4 k \paren {k + 1} + 1\)

But $k$ and $k + 1$ are of opposite parity and can therefore be expressed as $2 r$ and $2 s + 1$ (either way round).

\(\ds x\)

\(=\)

\(\ds 4 k \paren {k + 1} + 1\)

\(\ds \)

\(=\)

\(\ds 4 \paren {2 r} \paren {2 s + 1} + 1\)

\(\ds \)

\(=\)

\(\ds 8 r \paren {2 s + 1} + 1\)

Hence the result.

$\blacksquare$

Examples

$49$ Modulo $8$

\(\ds 49\)

\(=\)

\(\ds 7^2\)

\(\ds \)

\(=\)

\(\ds \paren {6 \times 8} + 1\)

$169$ Modulo $8$

\(\ds 169\)

\(=\)

\(\ds 13^2\)

\(\ds \)

\(=\)

\(\ds \paren {21 \times 8} + 1\)

Also see

• Square Modulo 8

Sources

• 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.1$ The Division Algorithm