A number is a perfect square if and only if it has odd number of positive divisors I believe I have the solution to this problem but post it anyway to get feedback and alternate solutions/angles for it.
For all $n \in \mathrm {Z_+}$ prove $n$ is a perfect square if and only if $n$ has odd # of positive divisors.
$\Rightarrow$: If $n$ is a perfect square it must consist of 1+ prime factors each to an even power.  If $n$ consists of 1+ prime factors each to an even power it must have an odd # of positive divisors because each even power contributes itself plus $1$(for the $0$ case) to the number of divisors for product $n$.
$\Leftarrow$: If $n$ has an odd number of positive divisors it must consist of 1+ prime factors each to an even power.  If $n$ consists of 1+ prime factors each to an even power it must be a perfect square.
Thanks.

Your forward direction looks good. The accepted answer does a good job using the conventional notation for the sum of positive divisors, $\nu(n)$. Here's the reverse direction.
Let $n$ be some square. Now, $n^2=n\cdot n=(p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_l^{\alpha_l})\cdot(p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_l^{\alpha_l})=p_1^{2\alpha_1}p_2^{2\alpha_2}\cdots p_l^{2\alpha_l}$. It is clear the exponents (meaning the $2\alpha_i$'s) of the prime factorization must be even. Thus, $\nu(n^2)=(2\alpha_1+1)(2\alpha_2+1)\cdots(2\alpha_l+1)$. Evidently each term of $\nu(n^2)$ must be odd, which means their product will be odd, i.e., $\nu(n^2)$ is odd.
A worked example is $\nu(196)$: $196=14\cdot14=(2^1\cdot7^1)\cdot(2^1\cdot7^1)=2^{2(1)}\cdot7^{2(1)}=2^2\cdot7^2$. $\nu(196)=\nu(14^2)=(2+1)(2+1)=9$. Evidently $196$ has $9$ positive divisors: $1, 2, 4, 7, 14, 28, 49, 98, 196$.
