Show that for every $(a, b, c) \in \mathbb{Z}^3$ with $a^2 + b^2 = c^2$ $60 \mid xyz$

Question

How would I approach this problem?

Let $(a, b, c) \in \mathbb{Z^3}$ with $a^2 + b^2 = c^2$. Show that: $$ 60 \,\mid\, abc $$

Answer 1

If one of them is 0, then the product is 0, divisible by 60.

Assume that $abc\neq 0$, WLOG $a,b,c>0$.

Then $a=m^2-n^2$, $b=2mn$, $c=m^2+n^2$ for some $m,n\in\mathbb{N}$ (well known thing, https://en.wikipedia.org/wiki/Pythagorean_triple), so $abc=2mn(m^2-n^2)(m^2+n^2)$.

$60=4\cdot 3\cdot 5$ and $4,3,5$ relatively prime.

If $2|m$ or $2|n$, then $4|2mn$; else $2|m^2-n^2$ and $2|m^2+n^2$, so $4|(m^2-n^2)(m^2+n^2)$; hence the divisibility by 4 is proved.

If $3|m$ or $3|n$, then $3|mn$; else $m^2\equiv 1\equiv n^2\pmod{3}$, so $3|m^2-n^2$ and the divisibility by 3 is proved.

If $5|m$ or $5|n$, then $5|mn$; else $m^2$ and $n^2$, as squares, are 1 or 4 mod 5. If $m^2\equiv n^2\pmod{5}$, then $5|m^2-n^2$, if not, then $5|m^2+n^2$, since in this case $m^2+n^2\equiv 1+4$. This gives the divisibility by 5.

Answer 2

There are formulas for the integer solutions of $a^2+b^2=c^2$: you can use them to give a proof.

If you don't know these formulas, here is an alternative method using modular arithmetic.

First we show that $5\mid abc$. This is clearly true if $5\mid a$ or $5\mid b$; now assume that neither of these is the case. The squares modulo $5$ are $0,1,4$; since $a,b\not\equiv0$ we have $$c^2=a^2+b^2\equiv\langle1\ \hbox{or}\ 4\rangle+\langle1\ \hbox{or}\ 4\rangle\equiv0\ \hbox{or}\ 2\ \hbox{or}\ 3\pmod5\ .$$ But $2,3$ are impossible, so $c^2\equiv0$, so $5\mid c$, so $5\mid abc$.

Now show that $3\mid abc$ by using a similar method. In fact, we have $3\mid a$ or $3\mid b$, because if neither of these is the case then $$c^2=a^2+b^2\equiv1+1\equiv2\pmod3\ ,$$ which is impossible. Hence $3\mid abc$.

Now consider the situation modulo $8$, in which the squares are $0,1,4$. If $a,b$ are both odd then $$c^2=a^2+b^2\equiv1+1\equiv2\pmod8$$ which is impossible. If $a,b$ are both even then obviously $4\mid abc$. If $a$ is even and $b$ is odd then $c$ is odd and $$a^2=c^2-b^2\equiv1-1\equiv0\pmod8\quad\Rightarrow\quad 8\mid a^2\quad \Rightarrow\quad 4\mid a$$ and so $4\mid abc$; similarly, if $b$ is even and $a$ is odd then $4\mid abc$.

We have shown that $3,4,5\mid abc$; and $3,4,5$ are coprime in pairs; so $60\mid abc$.

Answer 3

\begin{align} n^2 \mod 3 &\equiv 0 \text{ or } 1\\ a^2 + b^2 &\equiv c^2 \mod 3 \quad\text{the following two cases are the only possibilities}\\ 0+0 &\equiv 0 \mod 3 \\ 0+1 &\equiv 1 \mod 3\\ \end{align} Regardless, at least one number in the triple is divisible by $3$.

\begin{align} n^2 \mod 5 &\equiv 0 \text{ or } 1 \text{ or } 4\\ a^2 + b^2 &\equiv c^2 \mod 5 \quad\text{the following four cases are the only possibilities}\\ 0+0 &\equiv 0 \mod 5 \\ 0+1 &\equiv 1 \mod 5\\ 0+4 &\equiv 4 \mod 5\\ 4+1 &\equiv 0 \mod 5\\ \end{align} Regardless, at least one number in the triple is divisible by $5$.

$4\big|abc $ if there are at least $2$ twos in all three. Note that even if $n^2$ is divisible by $4$, then it may not have a factor of $4$ because it could be of the form $2 \cdot p_1^{q_1}\cdot p_2^{q_2}...p_n^{q_n}$, which is only divisible by $2$. Therefore, we must check if $n^2$ has a $16$. However, we can use $8$ instead, because if $n^2$ has an $8$, it must have a $16$ (because $8$ is not a perfect square.)

\begin{align} \text{if } n &\equiv 1 \mod 2\\ n^2 \mod 8 &\equiv 1 \\ \text{if } n &\equiv 0 \mod 2\\ n^2 \mod 8 &\equiv 0 \text{ or } 4 \\ a^2 + b^2 &\equiv c^2 \mod 5 \quad\text{the following four cases are the only possibilities}\\ 0+0 &\equiv 0 \mod 8 \\ 0+1 &\equiv 1 \mod 8\\ 0+4 &\equiv 4 \mod 8\\ 4+4 &\equiv 0 \mod 8\\ \end{align}

Now we can see that the product $abc$ has at least $3\cdot 4 \cdot 5$, which are the factors of $60$