9

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.

Tortar
  • 3,980
  • okay what do you know about sixth powers of odd numbers ? –  Feb 04 '20 at 20:57
  • 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. –  Feb 04 '20 at 21:05
  • ok , I agree with you that it has an odd number of divisors because it's a square , but why a multiple of 7 ? – Tortar Feb 04 '20 at 21:08
  • 1
    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. –  Feb 04 '20 at 21:17
  • I don't think you need to worry about putting in terms of $2k+1$. I think it is enough to assume that $n$ is not divisible by $2$ and all its divisors are odd. – fleablood Feb 04 '20 at 22:11
  • use the the number of divisors is of form $6j+1$ ? –  Feb 04 '20 at 23:40
  • Wow... I don't have it. but if $n=p^k$ for a prime $p$ then the sum of the factors is $1+p+ ..... + p^{k}=\frac {p^{k+1}-1}{p-1}$ which is relatively prime to $n^6$. To generalize if $n=\prod p_i^{k_i}$ then the sum of the factors of $n$ is $\prod\frac {p_i^{k_i+1}-1}{p_i-1}$ ... oh, wait. The questin is the sum of divisors of $n^2$ do not divide $n^6$.....Hmmm – fleablood Feb 05 '20 at 16:43
  • @Tortar If you could prove the case that $n$ is not a perfect square, it might help to post it. Moreover, upto which $k$ did you check this conjecture ? – Peter Feb 06 '20 at 10:00
  • The conjecture holds upto $\ k=10^7\ $. – Peter Feb 06 '20 at 10:17
  • Do you mean that it works at least until $k=10^7$ or that it is disproven there? @Peter – Tortar Feb 06 '20 at 11:31
  • @Tortar At least $k=10^7$, I currently check the range $10^7-10^8$ – Peter Feb 06 '20 at 11:33
  • 1
    New range finished, so upto $k=10^8$, the conjecture still holds. – Peter Feb 06 '20 at 11:34
  • What is the significance of $n^3$? Are there counterexamples for $n^2$ or even $n$? – Mees de Vries Feb 06 '20 at 12:42
  • 1
    @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. – Ivan Neretin Feb 06 '20 at 12:44
  • Oops, my bad. What about counterexamples for $n^4$, or $n^k$ for any $k$? – Mees de Vries Feb 06 '20 at 12:45
  • I have left $n^3$ just because I found this conjucture initially with this value . – Tortar Feb 06 '20 at 12:54

1 Answers1

3

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.

Mees de Vries
  • 26,947
  • So your point is that should be more probable to find a counterexample with a not composite odd number ($2k+1$)? – Tortar Feb 06 '20 at 13:54
  • 1
    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. – Mees de Vries Feb 06 '20 at 13:55
  • 1
    What strikes me here is that this argument is very similar to that used in heuristics for odd perfect numbers. – nickgard Feb 06 '20 at 14:10
  • @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. – ChinG May 11 '20 at 18:35
  • @ChinG I don't think that this is a definitive answer , I upvoted it anyway – Tortar May 11 '20 at 18:46