Prove that if $x$ is odd, then $x^2$ is odd 
Prove that if $x$ is odd, then $x^2$ is odd

Suppose $x$ is odd. Dividing $x^2$ by 2, we get:
$$\frac{x^2}{2} = x \cdot \frac{x}{2}$$
$\frac{x}{2}$ can be rewritten as $\frac{x}{2} = a + 0.5$ where $a \in \mathbb Z$. Now, $x\cdot\frac{x}{2}$ can be rewritten as:
$$x\cdot\frac{x}{2} = x(a+0.5) = xa + \frac{x}{2}$$
$xa \in \mathbb Z$ and $\frac{x}{2} \notin \mathbb Z$, hence $xa + \frac{x}{2}$ is not a integer. And since $xa + \frac{x}{2} = \frac{x^2}{2}$, it follows that $x^2$ is not divisible by two, and thus $x^2$ is odd.
Is it correct?
 A: Your written proof is correct. FYI, here is another proof technique which, although it's far more than you need in your case as Bill's comment indicates, is somewhat shorter and, perhaps, of some use to you, such as for other more complicated related problems.
Since $x$ is odd, this means it has no factors of $2$. By the Fundamental theorem of arithmetic, $x^2$ has the same prime factors as $x$, just twice as many of each of them and, thus, also no factor of $2$. As such, it is also odd.
A: Let $x = 2k + 1 \in \mathbb{Z}$ be odd. So, $$ x^{2} = (2k + 1)^2 = 4k^{2} + 4k + 1 = 2(2k^{2} + 2k) + 1$$ Let $t = 2k^{2} + 2k  \in \mathbb{Z}$, thus: $ x^{2} = 2t + 1$.
A: Your proof is correct.  
You could also prove by contrapositive:  
if $x^2$ is even, then $2|x^2=x\cdot x,$ so by Euclid's lemma $2|x,$ so $x$ is even.  
If you don't like that, you could say $x$ is odd, and the product of two odd numbers is odd, 
so $x^2=x\cdot x$ is odd.
If you don't like that, you could argue as follows:
if $x$ is odd, then $x-1=2k$ and $x+1=2k+2$ with $k \in\Bbb Z$, 
so $x^2=(x-1)(x+1)+1=(2k)(2k+2)+1=2(k(2k+2))+1$ is odd.
A: Yes, $ $ your proof is correct. $ $ You have $\ a = \color{#c00}{(x\!-\!1)/2}\in\Bbb Z\ $ so eliminating $\,a\,$ the proof becomes
$$\begin{align} &\dfrac{x^2}2 - \dfrac{x}2\, =\, \dfrac{x(\color{#c00}{x\!-\!1})}{\color{#c00}2}\in \Bbb Z\\[.2em]
\Rightarrow\ \ \ &\dfrac{x^2}2\in\Bbb Z\iff \dfrac{x}2\in\Bbb Z\\[.2em]
\Rightarrow\ \ \ & \ \ \ \ 2\nmid {x^2}\iff\ \  2\nmid x\\[.2em]
{\rm i.e.}\  \ \ & \ x^2\  {\rm odd}\ \iff x\ \  {\rm odd}
\end{align}\qquad\qquad\qquad\qquad$$
