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I wish to prove that square of a number is even if and only if the number itself is even. Let's call that number $a$, we need to prove that $$ a^2\text{ is even}\Leftrightarrow a\text{ is even} $$

We need to prove two things:

  1. If $a$ is even, then $a^2$ is also even.

I think that I can do this part. If $a$ is even, then we can write $a=2k$ where $k$ is some integer. Now $$ a^2=(2k)^2=4k^2=2(2k^2) $$ which shows that $a^2$ is also even.

  1. If $a^2$ is even, then $a$ is also even.

I am having difficulty here. If $a^2$ is even, then I can write $a^2=2m$ . for some integer $m$ but taking square root gives $a=\sqrt{2m}$ which can be written as $a=2\sqrt{\frac{m}{2}}$ but the problem is that will $\sqrt{\frac{m}{2}}$ always end up as some integer? I do not think so!

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    The trick is to use the contrapositive. – 2017-01-25

3 Answers 3

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For part 2, suppose that $a$ is not even. Then $a$ is of the form \begin{align*} a = 2k + 1 \end{align*} Thus, $a^2$ is of the form \begin{align*} a^2 = 4k^2 + 4k + 1 \end{align*} Since $4k^2 + 4k$ is even, $4k^2 + 4k + 1$ must be odd. Thus $a^2$ is odd, a contradiction.

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Hint for the $2^{nd}$ part: by Euclid's lemma if a prime divides a product, then it must divide (at least) one of the factors, so $2 \mid a^2 = a \cdot a \implies 2 \mid a\,$.

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Possible generalization is as follows :

Let $n$ be a positive integer et let $p$ be a prime number. Then $p\mid n$ if, and only if, $p\mid n^2$.

One way is obvious. The other one may be seen as a direct consequence of Euclid's lemma.