Divisibility of Elements of Pythagorean Triple by 7

Theorem
Let $\left({a, b, c}\right)$ be a Pythagorean triple such that $a^2 + b^2 = c^2$.

Then at least one of $a$, $b$, $a + b$ or $a - b$ is divisible by $7$.

Proof
It is sufficient to consider primitive Pythagorean triples.

From Solutions of Pythagorean Equation, the set of all Pythagorean triples is generated by:
 * $\left({2 m n, m^2 - n^2, m^2 + n^2}\right)$

where:
 * $m, n \in \Z_{>0}$ are (strictly) positive integers
 * $m \perp n$, that is, $m$ and $n$ are coprime
 * $m$ and $n$ are of opposite parity
 * $m > n$.

Thus let:
 * $a := 2 m n$
 * $b := m^2 - n^2$

The proof proceeds by examining all possibilities of the congruences modulo $7$ of $m$ and $n$ subject to the constraint that:
 * $m$ and $n$ are of opposite parity.

The following results are used throughout:

From Modulo Addition is Well-Defined:
 * If $r \equiv s \pmod 7$ and $x \equiv y \pmod 7$, then:
 * $r + x \equiv s + y \pmod 7$
 * $r - x \equiv s - y \pmod 7$

From Congruence of Powers:
 * $r \equiv s \pmod 7 \implies r^2 \equiv s^2 \pmod 7$

First we dispose of the case that either:
 * $m \equiv 0 \pmod 7$

or:
 * $n \equiv 0 \pmod 7$

Then:
 * $2 m n \equiv 0 \pmod 7$

and so $a$ is divisible by $7$.

The remainder of the cases will be presented conveniently in tabular form.

Recall that:
 * $a := 2 m n$
 * $b := m^2 - n^2$

and so:
 * $a \bmod 7 = 2 m n \bmod 7$
 * $b \bmod 7 = \left({m^2 - n^2}\right) \bmod 7$

All columns of the below are understood to be integers modulo $7$:


 * $\begin{array} {c c | c c | c c c c | l l}

m & n & m^2 & n^2 & 2 m n = a & m^2 - n^2 = b & \left({a - b}\right) & \left({a + b}\right) & \text{Example}\\ \hline 1 & 2 & 1 & 4 & 4 & -3 \equiv 4 & 0 & 1 & \left({60, 221, 229}\right): & 221 - 60 = 7 \times 23 \\ 1 & 4 & 1 & 2 & 1 & -1 \equiv 6 & -5 \equiv 2 & 0 & \left({120, 209, 241}\right): & 120 + 209 = 7 \times 47 \\ 1 & 6 & 1 & 1 & 5 & 0 & 5 & 5 & \left({180, 189, 261}\right): & 189 = 7 \times 27 \\ \hline 2 & 1 & 4 & 1 & 4 & 3 & 1 & 0 & \left({3, 4, 5}\right): & 3 + 4 = 7 \times 1 \\ 2 & 3 & 4 & 2 & 5 & 2 & 3 & 0 & \left({96, 247, 265}\right): & 96 + 247 = 7 \times 49 \\ 2 & 5 & 4 & 4 & 6 & 0 & 6 & 6 & \left({160, 231, 281}\right): & 231 = 7 \times 33 \\ \hline 3 & 2 & 2 & 4 & 5 & -2 \equiv 5 & 0 & 3 & \left({5, 12, 13}\right): & 12 - 5 = 7 \times 1 \\ 3 & 4 & 2 & 2 & 3 & 0 & 3 & 3 & \left({136, 273, 305}\right): & 273 = 7 \times 39 \\ 3 & 6 & 2 & 1 & 1 & 1 & 0 & 2 & \left({204, 253, 325}\right): & 253 - 204 = 7 \times 7 \\ \hline 4 & 1 & 2 & 1 & 1 & 1 & 0 & 2 & \left({8, 15, 17}\right): & 15 - 8 = 7 \times 1 \\ 4 & 3 & 2 & 2 & 3 & 0 & 3 & 3 & \left({7, 24, 25}\right): & 7 = 7 \times 1 \\ 4 & 5 & 2 & 4 & 5 & -2 \equiv 5 & 0 & 3 & \left({320, 999, 1049}\right): & 999 - 320 = 7 \times 97 \\ \hline 5 & 2 & 4 & 4 & 6 & 0 & 6 & 6 & \left({20, 21, 29}\right): & 21 = 7 \times 3 \\ 5 & 4 & 4 & 2 & 5 & 2 & 3 & 0 & \left({9, 40, 41}\right): & 9 + 40 = 7 \times 7 \\ 5 & 6 & 4 & 1 & 4 & 3 & 1 & 0 & \left({228, 325, 397}\right): & 228 + 325 = 7 \times 79 \\ \hline 6 & 1 & 1 & 1 & 5 & 0 & 5 & 5 & \left({12, 35, 37}\right): & 35 = 7 \times 5 \\ 6 & 3 & 1 & 2 & 1 & -1 \equiv 6 & -5 \equiv 2 & 0 & \left({120, 391, 409}\right): & 120 + 391 = 7 \times 73 \\ 6 & 5 & 1 & 4 & 4 & -3 \equiv 4 & 0 & 1 & \left({11, 60, 61}\right): & 60 - 1 = 7 \times 7 \\ \hline \end{array}$

Note that when the result has been demonstrated for $\left({m, n}\right)$, the result automatically also follows for $\left({n, m}\right)$. Hence the number of rows in the above could be halved.