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.

If two integers multiply to equal N, you will add two divisors to your total for that number.  This will be true for all values except when the two integers are the same.  Divisors will always be added in pairs except in this case.  So the only way to get to an odd total, is when the two divisors are equal, or N is a perfect square.
