Originally Posted by

**tomasklos** Well, of course if $\displaystyle x$ is odd, then $\displaystyle x^2$ is odd, and if $\displaystyle x$ is even, then $\displaystyle x^2$ is even. But without wanting to sound pedantic, I'd say that's a contrapositive proof in disguise. I'd like to *start* reasoning about $\displaystyle x^2$ and *end* at $\displaystyle x$.

I just discussed it with a colleague, and we came up with this.

By the fundamental theorem of arithmetic, $\displaystyle x^2$ has a unique prime factorization, so $\displaystyle x^2=p_1^{c_1}\cdot p_2^{c_2}\cdot\ldots\cdot p_n^{c_n}$, for certain primes $\displaystyle p_i$, and positive whole numbers $\displaystyle c_i$. (Assume these primes are listed in ascending order.) Because $\displaystyle x^2$ is even, $\displaystyle p_1=2$ and because it is a square, we have that for all $\displaystyle i$, $\displaystyle 2\mid c_i$. (This shows why, if $\displaystyle x^2$ is even, $\displaystyle 4\mid x^2$.) So $\displaystyle x=2^{c_1/2}\cdot p_2^{c_2/2}\cdot\ldots\cdot p_n^{c_n/2}$, which means $\displaystyle x$ is even. QED.