# Conjecture : an odd perfect square $n>1$ raised to the $m$-th power is never divisible by the sum of $n$'s divisors

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.

• okay what do you know about sixth powers of odd numbers ?
– user645636
Feb 4, 2020 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.
– user645636
Feb 4, 2020 at 21:05
• 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.
– user645636
Feb 4, 2020 at 21:17
• New range finished, so upto $k=10^8$, the conjecture still holds. Feb 6, 2020 at 11:34
• @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. Feb 6, 2020 at 12:44

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.
• So your point is that should be more probable to find a counterexample with a not composite odd number ($2k+1$)? Feb 6, 2020 at 13:54
• 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. Feb 6, 2020 at 13:55