Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I think the title is quite clear.


$$ n = \prod_{i=1}^n p_i ^{k_i}$$

is it possible to know something about the prime factorization of $n+1$? (I mean in terms of $p_i, k_i$)

share|cite|improve this question

marked as duplicate by Micah, Mike Miller, 900 sit-ups a day, Mathmo123, Will Jagy Aug 11 '14 at 0:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Well, $k_i \geq 1$ implies that $p_i$ does not divide $n+1$, since $\gcd (n,n+1)=1$. – vociferous_rutabaga Aug 10 '14 at 23:07
It's possible to factor $n+1$ if we know $n$. Because we know how to factor integers. It's just slow. – blue Aug 10 '14 at 23:40

If your definition of "something" is very weak, then yes. For example, none of the $p_i$ can appear in the prime factorization of $n+1$.

In some very simple cases, such as when $n$ is a perfect power, there may be algebraic identities that make your life easier. For example, if $n=k^3$, then $n+1=k^3+1=(k+1)(k^2-k+1)$. There's no guarantee that either of $k+1$ or $k^2-k+1$ is prime, but at least they're smaller than $n+1$: that is, you've gotten started...

Even in certain extraordinarily simple cases, though, it can be hard to get anywhere. For example, it's still an open problem whether $2^{2^k}+1$ is prime for infinitely many $k$ — and the numbers $2^{2^k}$ have about as simple a prime factorization as you could imagine!

share|cite|improve this answer

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