I was dealing with a different problem when this question struck me-

Let us have a natural number $n$, and we know that the divisors of $n$ are $d_1,d_2,\dots d_k$. Now, we add any other $x\in \mathbb N$ to $n$ and get a natural number $y=(x+n)$. What can be said about the divisors of $y$ (let's say $d_1^{\prime},d_2^{\prime}\dots d_{k^{\prime}}^{\prime}$). Can we relate the $d_i$'s with the $d_j^{\prime}$'s? If not, then what extra information do we need?

I searched for an answer online, but nothing significant popped out. When I started thinking about it, I realised that there are a few obvious answers like "$n$ and $x$ are of the same parity if and only if $y$ is divisible by a positive power of $2$" or "$y$ must be divisible by the greatest common divisor of $n$ and $x$". But, I can't think of anything more concrete. I also tried to use the divisor counting and the divisor sum function, but that didn't help either.

Also, the last line in the question (i.e., what extra information do we need) may be met with "we need to know the exact value of $x$". But, of course, I want to minimize the number of requirements.

So, can anybody help me to have a detailed study of this idea? I still can't believe that nothing about this is (easily) available. I'm sure, people must have thought about this before. So, if anybody is aware of any such researches, please inform.

Edit: After reading the comments, I wanted to specify some particularities-

1. I already mentioned that I want something more than the trivialties of if $d|y$, then $d|n \iff d|x$. user2661923 asked for the divisors of $x$ or the intersection of divisors of $n$ and $x$. We can assume that the intersection is empty- that's the case I'm most interested in. Even if we have an intersection, can we say anything more than "all elements of the intersection divides $y$"? As Arthur pointed out, even relating the divisors of $n$ and $n+1$ seems to be extremely difficult. Also, even if we take two primes $p_1$ and $p_2$ (so that we need to deal with the least number of divisors), can we say anything about the divisors of $p_1+p_2$?

2. About the question by RoddyMacPhee, what I mainly want is to find a relation between the divisors of $y$ and the divisors of $n$. If that's not directly possible, I want to know what are the minimal requirements to get to such a relation. Or is it that divisors are so faithful towards multiplication that addition just randomly shuffles them without any significant mathematical relation?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Pedro
    Jul 6, 2021 at 7:30

1 Answer 1


Question: "Let us have a natural number n, and we know that the divisors of n are d1,d2,…dk. Now, we add any other x∈N to n and get a natural number y=(x+n). What can be said about the divisors of y (let's say d′1,d′2…d′k′). Can we relate the di's with the d′j's? If not, then what extra information do we need?"

Answer: Let $n:=p_1^{l_1}\cdots p_k^{l_k}$ where the $p_i$ are distinct prime numbers and $l_i \geq 1$. Let $y:=q_1^{a_1}\cdots q_m^{a_m}$ where the $q_i$ are distinct prime numbers and $a_j\geq 1$ and with $x:=y-n >0$. It follows you may construct any such $y$ by adding a natural number $x$: There is by construction an equality $n+x=y$.

Example: As an example: $l=2, p_1:=2,p_2:=3$ and $n:=2^a3^b$. Let $3<q_1< \cdots <q_{10000}:=q_l$ be a set of 10000 distinct prime numbers larger than $3$ and let $y:=q_1^{a_1}\cdots q_l^{a_l}$ for some integers $a_i \geq 1$. The number $x:=y-n$ is a natural number and $y=x+n$. Is there a relation between $n$ and $y$? It seems not - you may choose arbitrary distinct prime numbers $q_i\notin \{2,3\}$, and the number $y$ will not be "determined" by the prime numbers $2,3$.

  • $\begingroup$ Nah the possibilities are set out by $n$ and $x$ as they'll have to be additive inverses modulo any divisors of $y$ $\endgroup$ Jul 5, 2021 at 15:50
  • $\begingroup$ @RoddyMacPhee - write a detailed post where you explain this. $\endgroup$
    – hm2020
    Jul 6, 2021 at 9:49
  • $\begingroup$ What detail it's just $n+x\equiv 0\pmod{d^{\prime}_j}$ now take away either of $n,x$ away from both sides. $\endgroup$ Jul 6, 2021 at 20:26

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