📂Number Theory

One of the Pythagorean Triples Must Be a Multiple of 3

Definition ¹

When natural numbers $a,b,c$ satisfy $a^2 + b^2 = c^2$, either $a$ or $b$ is a multiple of $3$.

Explanation

Among the Pythagorean triples, not only must one of them be even, but we can also say that at least one is a multiple of $3$.

Proof

For some natural number $n$, let us consider natural numbers divided into three cases according to their remainder $1, 2, 0$ when divided by $3$.

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Case 1. When the remainder is $1$ $$ \begin{align*} (3n+1)^2 &= 9 n^2 + 6n + 1 \\ =& 3( 3 n^2 + 2n) + 1 \end{align*} $$ Thus the remainder of the square is $1$.

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Case 2. When the remainder is $2$ $$ \begin{align*} (3n+2)^2 =& 9 n^2 + 12n + 4 \\ =& 3 ( 3 n^2 + 4n + 1 ) + 1 \end{align*} $$ Thus the remainder of the square is again $1$.

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Case 3. When the remainder is $0$

A multiple of $3$, even when squared, is still divisible by $3$.

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In other words, every square number $n^2$ has a remainder of either $1$ or $0$ when divided by $3$.

If we assume that neither $a$ nor $b$ is a multiple of $3$, then both $a^2$ and $b^2$ have a remainder of $1$. Therefore, the remainder of $c^2 = a^2 + b^2$ divided by $3$ becomes $2$. However, since we showed earlier that no square number can have a remainder of $2$ when divided by $3$, this is a contradiction. Hence, either $a$ or $b$ must be a multiple of $3$.

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