Prove that one of the sides of a Pythagorean triangle is a multiple of 3.

Question

Prove that if $\{a,b,c\}$ forms a Pythagorean triple, then at least one of the three numbers must be divisible by $3$. Need some help with this one.

Answer 1

Hint: A perfect square is either 0 or 1 (mod) 3. The worst case scenario is that all are 1 (mod) 3.

Answer 2

Consider the Pythagorean theorem: $a^2 + b^2 = c^2$.

Let $a^2 = 3m$ or $a^2 = 3m + 1$ (A perfect square can be expressed as a multiple of $3$ or the successor of a multiple of $3$).

Similarly, let $b^2 = 3k$ or $b^2 = 3k + 1$.

Then $c^2$ according to the Pythagorean theorem, is the sum of $a^2$ and $b^2$.

Let us assume that we take $a^2$ and $b^2$ as non-multiples of $3$.

Then $c^2 = 3m + 1 + 3k + 1 = 3(m+k)+2$. Put simply, it is the successor of the successor of a multiple of $3$. This is a contradiction since $c^2$ is a perfect square which can only be expressed as a multiple of $3$ or the successor of a multiple of $3$.

Therefore, one leg has to be a multiple of $3$.

S Sandeep