If $a,b,c \in \mathbb{Z}$ and $a^2+b^2=c^2$ show that $abc \equiv 0 \pmod{60}$.

Question

If $a,b,c \in \mathbb{Z}$ and $a^2+b^2=c^2$ show that $abc \equiv 0 \pmod{60}$.

I once read, on a number theory textbook - forget the title, in one of the problems list that all Pythagorean triplets when multiplied are divisible by 60

Answer 1

In $\mathbb{Z}/4$ the square are $0$ and $1$, we cannot have $a=b=1$ mod $4$, in this case $c^2=2$ mod $4$ contradiction.

In $\mathbb{Z}/3$ the square are $0$ and $1$, we cannot have $a=b=1$ mod $3$, since in this case $c^2=2$ mod $3$.

In $\mathbb{Z}/5$ the square are $0,1,4$, we cannot have $a=b=1$ mod $5$ and $a=b=4$ mod $5$, since in this case $c^2=2$ mod $5$, we deduce that $a=1, b=4$ mod $5$, $b=1, a=4$ mod $5$, this implies that $c^2=0$ mod $5$ and $c=0$ mod $5$,

This implies $3,4$ and $5$ are divsors of $abc$ which implies that $abc$ is divisible by $60$.

$a=0$ mod $5$ or $b=0$ mod $5$.

Answer 2

According to the prime factorization $60=2^2\cdot 3\cdot 5$ so it suffices to show $abc\equiv 0\pmod 4$, $\pmod 3$, and $\pmod 5$.

$abc\equiv 0\pmod 4$: If two or all three of the numbers are even, we are done. Hence assume at least two are odd. As the sum of odd numbers is even, this means that exactly two of $a,b,c$ are odd. Thess cannot be $a$ and $b$ as that would make $c^2\equiv 1+1\pmod 8$, which is absurd. Hence $c$ and wlog $a$ are odd. Then $b^2=c^2-a^2\equiv 0\pmod 8$, which means $4\mid b$ and we are done.

$abc\equiv 0\pmod 3$: If any of $a,b,c$ is a multiple of $3$, we are done. So assume otherwise. But then $a^2\equiv b^2\equiv c^2\equiv 1\pmod 3$, whereas $1+1\not\equiv 1\pmod 3$.

$abc\equiv 0\pmod 5$: If any of $a,b,c$ is a multiple of $5$, we are done. So assume otherwise. But then each of $a^2,b^2,c^2$ is $\equiv \pm1\pmod 5$, so $a^2+b^2\equiv$ one of $-2,0,2$, but $\not\equiv \pm1$.