# Divisibility of Elements of Pythagorean Triple by 7

## Theorem

Let $\tuple {a, b, c}$ 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:

$\tuple {2 m n, m^2 - n^2, m^2 + n^2}$

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 $n$.

The following results are used throughout:

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 $7$.

Next we note that if:

$m \equiv n \pmod 7$

then:

$m^2 - n^2 \equiv 0 \pmod 7$

and so $b$ 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:

$a \bmod 7 = 2 m n \bmod 7$
$b \bmod 7 = \paren {m^2 - n^2} \bmod 7$

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) & \paren {a + b} & \text{Example}\\ \hline 1 & 2 & 1 & 4 & 4 & -3 \equiv 4 & 0 & 1 & \tuple {60, 221, 229}: & 221 - 60 = 7 \times 23 \\ 1 & 3 & 1 & 2 & 6 & -1 \equiv 6 & 0 & 5 & \tuple {48, 55, 73}: & 55 - 48 = 7 \times 1 \\ 1 & 4 & 1 & 2 & 1 & -1 \equiv 6 & -5 \equiv 2 & 0 & \tuple {120, 209, 241}: & 120 + 209 = 7 \times 47 \\ 1 & 5 & 1 & 4 & 3 & -3 \equiv 4 & -1 \equiv 6 & 0 & \tuple {39, 80, 89}: & 39 + 80 = 7 \times 17 \\ 1 & 6 & 1 & 1 & 5 & 0 & 5 & 5 & \tuple {180, 189, 261}: & 189 = 7 \times 27 \\ \hline 2 & 1 & 4 & 1 & 4 & 3 & 1 & 0 & \tuple {3, 4, 5}: & 3 + 4 = 7 \times 1 \\ 2 & 3 & 4 & 2 & 5 & 2 & 3 & 0 & \tuple {96, 247, 265}: & 96 + 247 = 7 \times 49 \\ 2 & 4 & 4 & 2 & 2 & 2 & 0 & 4 & \tuple {65, 72, 97}: & 72 - 65 = 7 \times 1 \\ 2 & 5 & 4 & 4 & 6 & 0 & 6 & 6 & \tuple {160, 231, 281}: & 231 = 7 \times 33 \\ 2 & 6 & 4 & 1 & 3 & 3 & 0 & 6 & \tuple {276, 493, 565}: & 493 - 296 = 7 \times 31 \\ \hline 3 & 1 & 2 & 1 & 6 & 1 & 5 & 0 & \tuple {20, 99, 101}: & 20 + 99 = 7 \times 17 \\ 3 & 2 & 2 & 4 & 5 & -2 \equiv 5 & 0 & 3 & \tuple {5, 12, 13}: & 12 - 5 = 7 \times 1 \\ 3 & 4 & 2 & 2 & 3 & 0 & 3 & 3 & \tuple {136, 273, 305}: & 273 = 7 \times 39 \\ 3 & 5 & 2 & 4 & 2 & -2 \equiv 5 & -3 \equiv 4 & 0 & \tuple {240, 551, 601}: & 240 + 551 = 7 \times 113 \\ 3 & 6 & 2 & 1 & 1 & 1 & 0 & 2 & \tuple {204, 253, 325}: & 253 - 204 = 7 \times 7 \\ \hline 4 & 1 & 2 & 1 & 1 & 1 & 0 & 2 & \tuple {8, 15, 17}: & 15 - 8 = 7 \times 1 \\ 4 & 2 & 2 & 4 & 2 & -2 \equiv 5 & -3 \equiv 4 & 0 & \tuple {44, 117, 125}: & 44 + 117 = 7 \times 23 \\ 4 & 3 & 2 & 2 & 3 & 0 & 3 & 3 & \tuple {7, 24, 25}: & 7 = 7 \times 1 \\ 4 & 5 & 2 & 4 & 5 & -2 \equiv 5 & 0 & 3 & \tuple {180, 299, 349}: & 299 - 180 = 7 \times 17 \\ 4 & 6 & 2 & 1 & 6 & 1 & 5 & 0 & \tuple {85, 132, 157}: & 85 + 132 = 7 \times 31 \\ \hline 5 & 1 & 4 & 1 & 3 & 3 & 0 & 6 & \tuple {24, 143, 145}: & 143 - 24 = 7 \times 17 \\ 5 & 2 & 4 & 4 & 6 & 0 & 6 & 6 & \tuple {20, 21, 29}: & 21 = 7 \times 3 \\ 5 & 3 & 4 & 2 & 2 & 2 & 0 & 4 & \tuple {156, 667, 685}: & 667 - 156 = 7 \times 73 \\ 5 & 4 & 4 & 2 & 5 & 2 & 3 & 0 & \tuple {9, 40, 41}: & 9 + 40 = 7 \times 7 \\ 5 & 6 & 4 & 1 & 4 & 3 & 1 & 0 & \tuple {228, 325, 397}: & 228 + 325 = 7 \times 79 \\ \hline 6 & 1 & 1 & 1 & 5 & 0 & 5 & 5 & \tuple {12, 35, 37}: & 35 = 7 \times 5 \\ 6 & 2 & 1 & 4 & 3 & -3 \equiv 4 & -1 \equiv 6 & 0 & \tuple {52, 165, 173}: & 165 + 52 = 7 \times 31 \\ 6 & 3 & 1 & 2 & 1 & -1 \equiv 6 & -5 \equiv 2 & 0 & \tuple {120, 391, 409}: & 120 + 391 = 7 \times 73 \\ 6 & 4 & 1 & 2 & 6 & -1 \equiv 6 & 0 & 5 & \tuple {104, 153, 185}: & 152 - 104 = 7 \times 7 \\ 6 & 5 & 1 & 4 & 4 & -3 \equiv 4 & 0 & 1 & \tuple {11, 60, 61}: & 60 - 1 = 7 \times 7 \\ \hline \end{array}$

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

$\blacksquare$