This is a conjucture that I created :

Let $\,n = (2k+1)^2 \,\, $with $k\in \mathbb{N}$ and so $n>1$, and let $$\,\,A = \sum_{d \in \mathbb{N}; \ d|n} d.$$ Then $n^m$ is never divisible by $A$ for every $m \in \mathbb{N}$ .

I found a proof for the simpler case with $n$ odd but not a perfect square :

An odd number which is not a square has a even number of divisors all odd . So their sum is even but the number raised to the $m$-th power is odd.

So if the conjucture was true then the theorem would be true for all odd numbers greater than $1$.

However I have no idea how to proceed to prove it in the case of an odd perfect square.

It seems rather linked with perfect numbers.

  • $\begingroup$ okay what do you know about sixth powers of odd numbers ? $\endgroup$
    – user645636
    Feb 4, 2020 at 20:57
  • $\begingroup$ I meant in general. we know it has a multiple of 7 ( okay thinking of $2k+1$ prime) divisors, and an odd number of divisors... etc. $\endgroup$
    – user645636
    Feb 4, 2020 at 21:05
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    $\begingroup$ I was thinking of prime factorization, number of divisors of $n=p_1^{a_1}p_2^{a_2}\cdots p_m^{a_m}$ is the product of one more than each exponent. $\endgroup$
    – user645636
    Feb 4, 2020 at 21:17
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    $\begingroup$ New range finished, so upto $k=10^8$, the conjecture still holds. $\endgroup$
    – Peter
    Feb 6, 2020 at 11:34
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    $\begingroup$ @Mees Obviously, if there would be counterexamples for $n^2$ or $n$ (the latter is impossible, but whatever), they would be counterexamples for $n^3$ as well. $\endgroup$ Feb 6, 2020 at 12:44

1 Answer 1


Here is a heuristic argument (that is too long for a comment) for why you would not expect $A$ to divide $n^3$, or any $n^k$. If we consider the argument for $k$ arbitrary, what we are asking is: for odd $n > 1$, is it always the case that $\sigma(n)$ has a prime factor that is not a factor of $n$? As you've pointed out, if $n$ is non-square, then $\sigma(n)$ is even, so the conjecture is true, and this leaves the case when $n$ is a square.

If $n = p_1^{2k_1}p_2^{2k_2} \cdots p_l^{2k_l}$, then $$ \sigma(n) = \prod_{i = 1}^l \sigma(p_i^{2k_i}) = \prod_{i=1}^l \frac{p_i^{2k_i + 1} - 1}{p_i - 1}. $$ So, finding a counterexample comes down to the following problem: find a set of odd primes $\{p_1, \ldots, p_l\}$ with exponents $k_1, \ldots, k_l$ such that for each $i$, the number $$ \frac{p_i^{2k_i + 1} - 1}{p_i - 1} $$ factorizes into the primes $p_1, \ldots, p_l$. Now this number grows very fast (in $k$), and so it is unlikely to accidentally hit a fairly smooth number. Among the first 100 odd primes $p_i$, with $k_i$ ranging up to 10, there are only a few situations where this number is even $p_i$-smooth -- a necessity for the largest prime dividing $n$. This already shows that $n$ must be quite large, and this is in a sense a very weak argument.

  • $\begingroup$ So your point is that should be more probable to find a counterexample with a not composite odd number ($2k+1$)? $\endgroup$
    – Tortar
    Feb 6, 2020 at 13:54
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    $\begingroup$ No, if $n$ is a prime power then it's definitely not true, because then that prime doesn't divide $\sigma(n)$. You need $n$ to have at least two primes. But indeed a smoother $n$ (an $n$ with fewer distinct prime factors, the lower the better) is likely easier. $\endgroup$ Feb 6, 2020 at 13:55
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    $\begingroup$ What strikes me here is that this argument is very similar to that used in heuristics for odd perfect numbers. $\endgroup$
    – nickgard
    Feb 6, 2020 at 14:10
  • $\begingroup$ @Tortar , why don't you accept this answer? The respondent clearly put in time to respond to the answer, and the least you can do is accept it. $\endgroup$
    – ChinG
    May 11, 2020 at 18:35
  • $\begingroup$ @ChinG I don't think that this is a definitive answer , I upvoted it anyway $\endgroup$
    – Tortar
    May 11, 2020 at 18:46

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