# Given the prime factorization of $n$, is it possible to know anything about the prime factorization of $n+1$? [duplicate]

I think the title is quite clear.

Given

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

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

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

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!