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Define $\sigma(m) = \sum$ d : d|n. Prove that $p^j$$q^i$ cannot be a perfect number for $p, q$ odd, distinct primes.

Attempt at Solution: I have shown that $p^k$ can never be a perfect number, and im trying to use the multiplicative property of $\sigma$ to generalize to $p^j$$q^i$

Naz
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1 Answers1

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We have $$\sigma(p^jq^i)=\sigma(p^j)\sigma(q^i)=\frac{p^{j+1}-1}{p-1}\cdot \frac{q^{i+1}-1}{q-1}\lt \frac{p^{j+1}q^{i+1}}{(p-1)(q-1)}=p^jq^i \frac{p}{p-1}\cdot \frac{q}{q-1}.$$ Without loss of generality we can assume that $p\lt q$. Note that $\frac{p}{p-1}\le \frac{3}{2}$ and $\frac{q}{q-1}\le \frac{5}{4}$. It follows that $\frac{p}{p-1}\cdot \frac{q}{q-1}\le \frac{15}{8}\lt 2$. We conclude that $p^jq^i$ is deficient, and in particular cannot be perfect.

André Nicolas
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  • I've been fiddling around with the geometric series! But I couldn't see that the product was less than 15/8. Would you be able to clarify why that's true?

    p/p-1 is less than 3/2? Is it becauseas p approaches infinity, the fraction approaches 1? Thus the larger p is, the smaller the residue is?

     – Naz Feb 15 '15 at 14:46
  • Yes, you have described the reason. A more mechanical way of seeing it is to note that $\frac{p}{p-1}=\frac{p-1+1}{p-1}=1+\frac{1}{p-1}$, It is clear that $\frac{1}{p-1}$ decreases as $p$ increases. So the two largest values of $1+\frac{1}{p-1}$, for odd primes $p$, are at $p=3$ and $p=5$. – André Nicolas Feb 15 '15 at 16:31