Difference between Odd Squares is Divisible by 8

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Theorem

Let $a$ and $b$ be odd integers.

Then $a^2 - b^2$ is divisible by $8$.


Proof

Let $a = 2 m + 1$, $b = 2 n + 1$.

Then:

\(\ds a^2 - b^2\) \(=\) \(\ds \paren {2 m + 1}^2 - \paren {2 n + 1}^2\)
\(\ds \) \(=\) \(\ds \paren {4 m^2 + 4 m + 1} - \paren {4 n^2 + 4 n - 1}\)
\(\ds \) \(=\) \(\ds 4 \paren {m^2 - n^2} + 4 \paren {m - n}\)
\(\ds \) \(=\) \(\ds 4 \paren {m + n} \paren {m - n} + 4 \paren {m - n}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds 4 \paren {m + n + 1} \paren {m - n}\)


Suppose $m - n$ is even such that $m - n = 2 k$.

Then:

$a^2 - b^2 = 4 \paren {2 k} \paren {m + n + 1} = 8 k \paren {m + n + 1}$

and so is divisible by $8$.


Suppose $m - n$ is odd such that $m - n = 2 k + 1$.

Then:

$m + n + 1 = m + \paren {2 k + 1 + m} + 1 = 2 m + 2 k = 2 \paren {m + k}$

and so:

$a^2 - b^2 = 4 \paren {2 k + 1} 2 \paren {m + k} = 8 \paren {2 k + 1} \paren {m + k}$

and so is again divisible by $8$.

The result follows.

$\blacksquare$


Sources

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