Factorization special number

Can a number $$n=p_1\times p_2\times \dots\times p_k+1$$ where $$p_1, p_2,..., p_k$$ are primes, be easily factorized? (the $$k$$ primes are known)

• If there would be an easy way, then for example Collatz conjecture would be "easy" ... but at least you know $p_i \nmid n$. Btw I think you are using $n$ in two different meanings (number of primes and resulting number) – Sil Sep 22 at 11:57

If such a number could be easily factorised, then factorising $$n$$ would be as easy as factorising $$n-1$$. Which would be as easy as factorising $$n-2$$. And so on. So all you would have to do would be find a number $$n-k$$ that you can factorise, for some reasonably small $$k$$, and you could leapfrog your way to a factorisation of $$n$$.
• How do you obtain the factorization of $n+1$ from the factorization of $n$? – mip Sep 22 at 14:30
• @mip: That is equivalent to what your question is positing: given the factorisation $p_1\times p_2\times \dots\times p_k$ of an integer $m$, can we factorise $n = m+1$? – TonyK Sep 22 at 14:48
Not exactly. Until we know one of the primes is 2 or not, we can't tell if this value is even or odd. Until we know if 3 is one of the primes, or an odd number of the primes are 2 mod 3, we can't tell if 3 is a factor. Unless an odd number of primes are 3 mod 4, your value won't be divisible by 4. etc. But that's in effect trial division. Also it can be prime. $$2\cdot 3+1=7$$ as an example.