<s>[INST] Let $a$ and $b$ be elements of odd order in a group. Show that $a^2$ and $b^2$ commute if and only if $a$ and $b$ commute. I really don't know how to solve this problem. I just know that if $|a|=2k_1+1$ and $|b|=2k_2+1$, then, 
$a^{2k_1+1}=e=a^0$ and  $b^{2k_2+1}=e=b^0$. Also, if $|G|=n$, then, $2k_1+1,2k_2+1\equiv 0 (\mod n)$. 
 [/INST] You got it the other way round. It's $n \equiv 0 \pmod{2k_1 + 1}$ and $n \equiv 0 \pmod{2k_2 + 1}$.
Suppose $a^2b^2 = b^2a^2$. Then:
\begin{align*}
ab &= a^{2k_1+1}(ab)b^{2k_2+1} \\
&= a^{2k_1+2}b^{2k_2+2} \\
&= (a^2)^{k_1+1}(b^2)^{k_2+1} \\
&= (b^2)^{k_2+1}(a^2)^{k_1+1} \\
&= b^{2k_2+1}(ba)a^{2k_1+1} \\
&= ba
\end{align*}
</s>