Could a square be a perfect number?

Question

A perfect number is the sum of its (positive) divisors (excluding itself). I am wondering if a square could be a perfect number.

If it is an odd square, then, excluding itself, it has an even number of divisors which are odd. Adding them together yields an even sum. Therefore, an odd square could not be a perfect number.

What about the even square case? Every help is appreciated.

Answer 1

No.

The Euclid-Euler theorem states that any even perfect number $n$ (we don't know whether there are any odd ones) is of the form $$ n = 2^{k-1}(2^k - 1) $$ with $2^k - 1$ prime, and furthermore that any $n$ of that form is perfect (this last part is relatively easy to prove, but it is the former part you need). This is clearly not a square, since $2^k - 1$ is strictly larger than $2^{k-1}$ and prime.

Answer 2

A perfect square can be written on the form $n=(p_1^{\alpha_1}\cdots p_k^{\alpha_k})^2$ where $p_i$ are some prime numbers and $\alpha_i> 0$ are integers.

The sum of the divisors of $n$ is given by

$$\sigma(n) = \prod_{i=1}^k \sigma(p_i^{2\alpha_i}) = \prod_{i=1}^k(1+p_i + \ldots+p_i^{2\alpha_1})$$

The term $1+p_i+\ldots+p_i^{2\alpha_i}$ is an odd number for all primes $p_i$ and integers $\alpha_i> 0$ and since the product of odd numbers is odd we see that $\sigma(n)$ has to be odd. However if $n$ is a perfect number then (by definition) $\sigma(n) = 2n$ which is an even number so a square cannot be a perfect number.

Answer 3

In fact, to generalize on @winther's result, note that $\sigma (n)$ is odd iff $n$ is of the form $k^2$ or $2k^2$.

The proof of this can be found in this post.

Thus, since for a perfect number $\sigma(n)=2n \equiv 0 \pmod 2$, a perfect number can neither be of the form $k^2$ or $2k^2$.

Thus, any square-or any number that is twice a square-is not a perfect number.

Answer 4

There is no number that is perfect and square. To show that, consider n to be a perfect natural number. Then we have to case:

1. When n is even, then by Euler theorem, $n=2^{p - 1}({2^p - 1})$ , where ${2^p - 1}$ is Mersenne prime. Therefore n is not a square, because $\sqrt{n} = \sqrt{2^{p - 1}({2^p - 1})} $ which is irrational!
2. When n is odd, we have a theorem says that if n is an odd perfect square, then $ n=pa^2$, where a is an integer. To prove it consider the equation $\phi{(n)}=2n $ and when $\phi{(p^k)} $ is odd?, after that clearly n can not be square for the same argument above.