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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 $p_i, k_i$)

José D.
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    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
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    It's possible to factor $n+1$ if we know $n$. Because we know how to factor integers. It's just slow. – anon Aug 10 '14 at 23:40
  • There is a method, which under certain conditions, can factor $N+1$ or $N-1$ or $N-x$ or $N+x$ if the factorization of $N$ is known. The reverse uses the same method, that is knowing the factors of $N-x$...it is possible to factor $N$.
    https://math.stackexchange.com/questions/4810956/can-factoring-90-help-factor-91     – user25406 Nov 21 '23 at 11:41

1 Answers1

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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!

Micah
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