I would like to show that for any integer $n \geq 0$, $3|\sigma(3n+2)$, where $\sigma(n)$ denotes the sum-of-divisors function.

I have been able to show some results, for example that if $(3n+2)$ has a prime factor of the form $6k-1$ which occurs with an odd exponent, then $3|\sigma(3n+2)$. However, this leaves some other cases too, and it seems like my approach is not the best.

Any suggestions?

Thanks in advance!

  • 1
    $\begingroup$ If $k \equiv 2 \pmod{3}$, then $k$ has at least one prime factor $p \equiv 2 \pmod{3}$ that divides $k$ with an odd power. $\endgroup$ – Daniel Fischer Aug 18 '13 at 21:33
  • $\begingroup$ @DanielFischer: Thanks for your comment! It should say "a prime factor of the form 6k-1". $\endgroup$ – Alexandre Vandermonde Aug 18 '13 at 21:45
  • 1
    $\begingroup$ That isn't necessarily the case, $2$ is also a possibility. $\endgroup$ – Daniel Fischer Aug 18 '13 at 21:46
  • $\begingroup$ To unravel Daniel's first comment: the fact that $m\equiv 2$ mod $3$ necessarily implies there is a prime $p$ with $p\equiv2$ mod $3$ whose exponent in $m$ is odd. You can prove this by contradiction $-$ show that on the contrary assumption (all prime divisors $p\equiv-1$ have even exponent) $m$ must be $0$ or $1$ mod $3$. $\endgroup$ – anon Aug 18 '13 at 21:56
  • $\begingroup$ Thank you for clearing this up! I realize now that it suffices to consider the primes on the form $3k \pm 1$ and use the same technique as I used for the special case. $\endgroup$ – Alexandre Vandermonde Aug 18 '13 at 22:00

Briefly: $$2\sigma(3n+2) = \sum_{d\mid 3n+2} d + \sum_{d\mid 3n+2}\frac{3n+2}{d}= \sum_{d\mid 3n+2} \left(d+\frac{3n+2}{d}\right)$$

And just show that all the terms of this sum are divisible by $3$.

  • $\begingroup$ I should stop trying to write LaTeX from my iPad. $\endgroup$ – Thomas Andrews Aug 18 '13 at 21:57
  • $\begingroup$ Looks great now! Thanks for the hint, it is very neat. $\endgroup$ – Alexandre Vandermonde Aug 18 '13 at 22:11
  • $\begingroup$ Why all the terms of this sum are divisible by 3 ? $\endgroup$ – Pranasas Dec 18 '14 at 18:36
  • 1
    $\begingroup$ Because if $d\equiv 1\pmod 3$ then $\frac{3n+2}{d}\equiv 2\pmod 3$ and visa versa. @Pranasas $\endgroup$ – Thomas Andrews Dec 18 '14 at 18:38
  • 1
    $\begingroup$ Because $d\frac{3n+2}{d}=3n+2\equiv 2\pmod 3$, so if $d\equiv 1\pmod 3$... @Dominik $\endgroup$ – Thomas Andrews Jun 22 '15 at 18:44

Hint: Consider all pairs of numbers such that $a\times b = n$. Show that $3 \mid a+b$.

Hint: There is a 'special case' that you have to check for completeness. Use the fact that 2 is not a quadratic residue modulo 3.

  • $\begingroup$ What is this special case? With Thomas' motivation for your first hint, it seems to me that the method works for all $n \geq 0$. $\endgroup$ – Alexandre Vandermonde Aug 18 '13 at 22:09
  • $\begingroup$ @AlexandreVandermonde The special case is to check when $a=b$. But we can't have $a^2 = 3k+2$. $\endgroup$ – Calvin Lin Aug 18 '13 at 22:14
  • $\begingroup$ I'm sorry, but I still fail to see why Thomas' formula below would not be correct in the case $a=b$ (that is, $d=\sqrt{3n+2}$, right?). We still want to add one term for each of the sums in the middle equation. $\endgroup$ – Alexandre Vandermonde Aug 18 '13 at 22:22
  • $\begingroup$ @AlexandreVandermonde Thomas' solution is correct. He does work around the case of $a^2 = 3n+2$, by doing the counting twice. So he did account for it (though not explicitly). $\endgroup$ – Calvin Lin Aug 18 '13 at 22:27
  • $\begingroup$ Alright, thanks for clearing that up! $\endgroup$ – Alexandre Vandermonde Aug 18 '13 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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