Are there any "perfect" squares?

Suppose by contradiction that $n$ is a perfect number which is also a perfect square. We're going to prove that the sum of all divisors of $n$ is odd (all divisors, including $n$ , not just the proper ones). That will lead to a contradiction, because a number is perfect if and only if the sum of all its divisors is equal to $2n$ .

Note that to prove that the sum of all divisors of $n$ is odd it's enough to prove that the sum of all odd divisors of $n$ is odd. It's clear that, since $n$ is a perfect square, its largest odd divisor $m$ must also be a perfect square. Suppose that $m=k^2$ , for some $k$ . Now, every odd divisor $j$ of $n$ which is less than $k$ pairs up with an odd divisor of (n) which is greater than $k$ (namely, $\frac{k^2}{j}$ ). Counting this pairs together with $k$ we see than $n$ has an odd number of odd divisors. So the sum of the odd divisors of $n$ is odd. But then the sum of all divisors of $n$ is odd and this finishes the proof.

It's also true that there are no perfect numbers which are also perfect cubes. The proof of this is much harder and can be found in the following paper:

Luis H. Gallardo " On a remark of Makowski about perfect numbers ", Elem. Math. 65 (2010), 121--126.

Perfect numbers are given by 2^(p - 1)* (2^p - 1), where the second factor is a Mersenne prime; so there cannot be a perfect square, or cube for that matter. Ed Gray