Is the hypotenuse of a triangle ever divisible by three (for primitive Pythagorean triples)?

Solution 1:

The hypotenuse of a primitive triple is never divisible by $3$. For let $(x,y,z)$ be a primitive triple, and suppose $3$ divides $z$. Then $3$ cannot divide $x$ or $y$, else the triple would not be primitive.

It follows that $x^2$ and $y^2$ have remainder $1$ on division by $3$, which means $x^2+y^2$ has remainder $2$ on division by $3$.

Remark: A somewhat more elaborate argument shows that if $p$ is a prime of the form $4k+3$, then $p$ cannot divide the hypotenuse of a primitive triple.

Solution 2:

Not significantly different but I'd write this out using the modulus. That is, numbers can be $-1, 0, 1 (\mod 3)$ and thus the squares of them can only be $0, 1$ now $0+0 \equiv 0 (\mod 3)$ so then for a primitive 3-tuple this is not possible; also $1+1 \equiv 2 \equiv - 1(\mod 3)$ and $\nexists c \in \mathbb{N}|c^2 \equiv -1(\mod 3)$. So the only possibility is $a \equiv 0, b\equiv 1, c\equiv 1$ drawing both results in one go: exactly one side is divisible by three in a primitive triple and that side is a leg.

Edit: mod 4 the situation is not different, the squares are still 0 and 1 $(\mod 4)$ and so exactly one side is divisible by four in a primitive triple and that side is a leg.

And it so happens that the smallest set, is an example when these two are different: 3,4,5 and the second 5, 12, 13 is when these two are the same. Isn't mathematics beautiful?