Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that if $\sigma(n)=2n+1$ then $n$ is an odd perfect square.

(Here, $\sigma(n)$ is the sum of the positive divisors of $n$ including 1 and $n$ itself.)

As I know, this $n$ is a quasiperfect number, and I just proved that $n$ is a perfect square or $\frac{n}{2}$ is a perfect square.

share|improve this question
add comment

1 Answer 1

up vote 6 down vote accepted

I had seen this result before, but not the proof, so I looked up Cattaneo's original argument. It's in Italian. What follows is his proof with some details filled in.

First, if $\sigma(n) = 2n+1$ then $\sigma(n)$ is odd.

If $n = \prod_{i=1}^r p_i^{a_i}$ is the prime factorization of $n$, then we know that $$\sigma(n) = \prod_{i=1}^r (1 + p_i + p_i^2 + \cdots + p_i^{a_i}).$$ Since $\sigma(n)$ is odd, though, each factor $1 + p_i + p_i^2 + \cdots + p_i^{a_i}$ must also be odd.

For each odd prime factor $p_i$, then, there must be an odd number of terms in the sum $1 + p_i + p_i^2 + \cdots + p_i^{a_i}$. Thus $a_i$ must be even. This means that if $p_i$ is odd, $p_i^{a_i}$ is an odd perfect square. The product of odd perfect squares is another odd perfect square, and therefore $n = 2^s m^2$, where $m$ is odd.

Now we have $\sigma(n) = (2^{s+1}-1)M$, where $M = \prod_{p_i \text{ odd}} (1 + p_i + p_i^2 + \cdots + p_i^{a_i})$. Since $\sigma(n) = 2n+1$, \begin{align*} &(2^{s+1}-1)M = 2^{s+1}m^2 + 1 \\ \implies &(2^{s+1}-1)(M - m^2)-1 = m^2. \end{align*} This means that $-1$ is a quadratic residue for each prime factor of $2^{s+1}-1$. A consequence of the quadratic reciprocity theorem, though, is that $-1$ is a quadratic residue of an odd prime $p$ if and only if $p \equiv 1 \bmod 4$. Thus all prime factors of $2^{s+1}-1$ are congruent to $1 \bmod 4$. The product of numbers congruent to $1 \bmod 4$ is still congruent to $1 \bmod 4$, so $2^{s+1}-1 \equiv 1 \bmod 4$. However, if $s > 0$, then $2^{s+1}-1 \equiv 3 \bmod 4$. Thus $s$ must be $0$. Therefore, $n = m^2$, where $m$ is odd.

share|improve this answer
    
This strikes me as a rather challenging homework problem. –  Mike Spivey Jan 6 '13 at 3:52
    
Maybe so, but, as it is a homework problem, is it a good idea to give so detailed a response? –  Gerry Myerson Jan 6 '13 at 4:10
    
@Gerry: A fair question - and one that has been debated extensively on meta, as I'm sure you are aware. –  Mike Spivey Jan 6 '13 at 6:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.