Elementary proofs help I'm taking Principles of Mathematics this semester and I came across a problem, but I don't know whether my proof is valid or not. I was hoping you could help me out.  It goes like this: 
Let  $x$ be a natural number. Prove or disprove: $27|x^{2}\Rightarrow 9|x$. This is my proof:  
Assume $x^{2}$ is divisible by 27. Then $x^{2}=27k$, for some $k\in\mathbb{N}$.
Let $l$ be a natural number such that $k=3l^{2}$. Then, 
$x^{2}=27k\\ x^{2}=(9)(3)k\\ x^{2}=(9)(3)(3l^{2})\\x^{2}=81l^{2}\\ x^{}\ \ =9l$
Thus $x$ is divisible by 9 and the proposition is true.
 A: Hint: by the unique factorization theorem, $\,x=3^ay\,$ where $a \ge 0 $ is the largest power of $3$ that divides $\,x\,$ and $\,\gcd(y,3)=1\,$.
Then $27=3^3 \mid 3^{2a}y^2 = x^{2}\,$ implies that $2a \ge 3\,$, which in turn implies that $a \ge 2$.
A: Proof $1$ (using fundamental theorem of arithmetic):
If you may use this theorem, then you have $x^2=3^3\cdot k$. Then since $x^2$ is a perfect square, the power of $3$ on the right must be at least $4$ (it must be an even number and clearly isn't $0$ or $2$). Thus $3^4$ divides $x^2$, so $3^2$ divides $x$.

Proof $2$ (only uses modular arithmetic and some basic properties of division):
Let's prove the contrapositive: suppose that $9 {\not|} \ x$, and then show that $27\not |\ x^2$.
If $x\equiv 1\pmod 9$, then $x^2\equiv 1\pmod 9$. Thus $x^2$ is not a multiple of $9$ and so cannot be a multiple of $27$. Continuing,
$x\equiv 2\implies x^2\equiv 4 $ 
$x\equiv 3\implies x^2\equiv 0 $ 
$x\equiv 4\implies x^2\equiv 7 $ 
$x\equiv 5\implies x^2\equiv 7 $ 
$x\equiv 6\implies x^2\equiv 0 $ 
$x\equiv 7\implies x^2\equiv 4 $ 
$x\equiv 8\implies x^2\equiv 1 $.
This leaves us with the possibilities that $x\equiv 3$ or $6$ mod $9$. If $x=9a+3$, then $x^2=81a^2+54a+9$. Since $27$ divides $81$ and $54$, we have $27|9$, a contradiction. Similarly, if $x=9b+6$, then $x^2=81b^2+108b+36$. Again, $27|36$ gives a contradiction.
