TITLE: Prime factorization of n+1
QUESTION [12 upvotes]: If $n=\prod_{i=1}^{k} p_i^{e_i}$ is a prime factorization of integer $n$.
Is there a quick way to find the prime factorization of $n+1$?
Or the only way to do it is recalculating the whole factorization?
Any references and/or articles on this problem?

REPLY [4 votes]: There are two questions asked:
Q1. Does the prime factorization of $n$ give a quick way to find the prime factorization of $n+1$? 
A1.No it does not (as noted in other answers). Knowing $n$ is prime is of no use.
Q2. Is it ever of any use or do you always just have to start from scratch? (as I will choose the rephrase the question.) 
A2. It is sometimes of use but not usually. When $n$ is a power of a smaller number there may be some help. Since the case of Fermat numbers was raised I will comment that factoring numbers of the form $2^e+1$ is somewhat easier (relative to the size) than numbers of the form $2^e+3$. Given $n=2^e$ we know that $2^{e/f}+1$ is a factor of $n+1$ where $f$ is any odd divisor of $e$. So that is a start to factoring $n+1$. In the case $n=2^{2^e},$ $n+1$ might be prime. To test primality one has Pépin's test. The numbers involved are so huge that this is not practical very far out. ALSO it is known that any candidate factor must be of the form $k2^e+1$ so in attacking $2^{2^{20}}+1$ (known composite, no factors known) we have already ruled out 99.9999% of the possible factors.