📂Number Theory

One of the Pythagorean triples must be a multiple of three.

Definition ¹

A natural number $a,b,c$ that satisfies $a^2 + b^2 = c^2$, then either $a$ or $b$ is a multiple of $3$.

Explanation

Among the Pythagorean triples, not only is one of them always even, but at least one is also a multiple of $3$.

Proof

Consider a natural number $n$ and divide it into three cases based on the remainder $1, 2, 0$ when divided by $3$.

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Case 1. 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 number is $1$.

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Case 2. 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 number is also $1$.

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Case 3. The remainder is $0$

When a multiple of $3$ is squared, it still divides evenly by $3$.

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Thus, all square numbers $n^2$ have a remainder of $1$ or $0$ when divided by $3$.

If we assume both $a$ and $b$ are not multiples of $3$, then both $a^2$ and $b^2$ have a remainder of $1$. Therefore, the remainder of $c^2 = a^2 + b^2$ when divided by $3$ is $2$. However, as shown earlier, all square numbers cannot have a remainder of $2$ when divided by $3$, which is a contradiction. Therefore, either $a$ or $b$ must be a multiple of $3$.

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