Odd Square Modulo 8
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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
Sources
- 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.1$ The Division Algorithm