Is it true always that if for every $a,b,c\in \mathbb{N}$ and $a^2+b^2=c^2$, such that if $3|c$, then $abc$ is a multiple of $3^4$

Question

If $3|c\implies c=3k$ for some $k\in \mathbb{N}$

Thus if $a^2+b^2=c^2\implies a^2+b^2=9k^2$

So suppose we have $a=8,b=15,c=17$

Thus here $3 \nmid{17}$ but if $17\times 3=51$ is taken, we are done.

So let $c=51$, then $a=24,b=45$

Now thus this so becomes that $a^2+b^2=c^2\implies a^2+b^2=9k^2 \implies 9k_1^2+9k_2^2=9k^2$

Thus $k_1^2+k_2^2=k^2$

Now if $\Pi abc$ is performed, we see that it is definitely a multiple of $3^3$

But we shall also see that even after stripping off that 3 from $a$ and $b$, we get $a=k_1,b=k_2$,

The rest is to prove that one of $k_1$ or $k_2$ is a multiple of 3

Answer 1

Lemma: If $a^2+b^2=c^2$ then at least one of integers $a,b,c$ is divisible by $3$.

Proof: If $3\mid c$ we are done. Say $3$ doesn't divide $c$. Then $$a^2+b^2\equiv 1 \pmod 3$$

So if $3$ doesn't divide none of $a$ and $b$ we have $$2\equiv 1 \pmod 3$$ a contradiction.

Answer 2

Modulo $3$ only $0$ and $1$ are squares, hence $x^2+y^2=0$ mod $3$ implies $x=y=0$ mod $3$. It follows that all three of $a$, $b$, $c$ are divisible by $3$. Canceling this common factor we obtain $a'^2+b'^2=c'^2$ which is only possible if at least one of $a'$, $b'$, $c'$ is $=0$ mod $3$.