Largest integer $n$ such that $3^n$ divides every $abc$ with $a$, $b$, $c$ positive integers, $a^2+b^2=c^2$, and $3|c$ 
Let $P$ denote the set { $abc$ : $a$, $b$, $c$ positive integers, $a^2+b^2=c^2$, and $3|c$}. What is the largest integer $n$ such that $3^n$ divides every element of $P$?

I first saw that as  $3|c\implies c=3k$ where $k\in \mathbb{N}$
$\implies a^2+b^2=9k^2$
Now this means $a^2\equiv r1(mod \space 9)$  & $b^2\equiv r2(mod  \space 9)$
where $r1+r2=9k_1$ where $k_1\in \mathbb{N}$
In order to get the largest integer $n$ s.t $3^n|\space \text{every element of P}$
We have to assume $a=3k$ and $b=3k$
Thus $abc=3^3\times(some \space number \ne a \space multiple \space \space of \space 3)$
Thus $n \space should \space be =3$
But the answer here is 4
I can't figure out the solution.Please help.
P.S. Is there any geometrical angle in this problem meaning can this $a^2+b^2=c^2$ be considered a circle somehow and proceed? I can't think of that approach.
OK here we have a solution, lets look at it:
As @ChristianF suggested as an answer to another of my question,
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.
Also suggested by @Christian Blatter 
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 $3$.
We will follow along the lines:
If $3|c \implies c=3k$ for some $k \in \mathbb{N}$
Now this means according to @Christian Blatter,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 $3$. 
Hence $a=3k_1,b=3k_2$ for some $k_1,k_2 \in \mathbb{N}$
Now we get $9k_1^2+9k_2^2=9k^2$
Cancelling the 9 from the above from LHS and RHS we get
$k_1^2+k_2^2=k^2$
which is analogous to the fact that 
we obtain $a'^2+b'^2=c'^2$ which is only possible if at least one of $a'$, $b'$, $c'$ is $=0$ mod $3$
Thus one of $k_1,k_2$ is still a multiple of 3
$\implies$ $a\times b \times c= 3^4 \times K$
Hence for $n=4$, which is the largest integer, $3^n|a\times b \times c$
Thanks to @ChristianF and @Christian Blatter for the insight

First, convince yourself that every solution of $a^2+b^2=c^2$ has $abc$ a multiple of 3. 
Then convince yourself that every solution with $c$ a multiple of 3 must be such that $(a/3)^2+(b/3)^2=(c/3)^2$ with $a,b,c$ all integers (I think you've already done this, so the first sentence above is all you really need). 
