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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.

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    $x^2|27$ means $27$ is divisible by $x^2$. The OP has this backwards. – 2017-02-16
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    How do you know that there exists $l$ so that $k =3l^2$? – 2017-02-16
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    Fixed it, sorry! – 2017-02-16
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    @JoshuaRuiter You're right, I don't :/ I got it from proving the converse, but I know that doesn't mean anything. What should I do? – 2017-02-16

2 Answers 2

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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 $a \ge 2$.

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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.