Is there a typo? I don't see why the term $h+l$ matters - it now means just a multiple of $k(k+2)(k+3)$...

VicKmath7 wrote: Is there a typo? I don't see why the term $h+l$ matters - it now means just a multiple of $k(k+2)(k+3)$... I will check it, but I'm sure that is not $k(k+2)(k+3).$

Hopeooooo wrote: VicKmath7 wrote: Is there a typo? I don't see why the term $h+l$ matters - it now means just a multiple of $k(k+2)(k+3)$... I will check it, but I'm sure that is not $k(k+2)(k+3).$ I checked it, it was $k(k+1)(k+2)(k+3).$

This is wrong, ignore.

k=2*(x^2-2)^4-1 where x is a positive integer, done.

Some motivation behind my example: $\pm2$ is quadratic nonresidue modulo $p=8t+5$, so one can easily find infinitely many triples $(2s^2-1,2s^2,2s^2+1)$, where no one number is divisible by any prime $p$ of kind $8t+5$. We need to do something with $2s^2-2$. It is only need that number $(s-1)s(s+1)$ is not divisible by any prime $p$ of kind $8t+5$, so we could apply an induction argument or construct an example directly: $k=8x^4-2, k+1=8x^4-1, k+2 = 8x^4, k+3=8x^4+1$ for some positive ineger $x$ that not divisible by any prime $p$ of kind $8t+5$.

VicKmath7 wrote: the last should follow from $\sum \frac {1}{p_i}< \frac {1}{4}$. Unfortunately it is wrong, this sum tends to infinity.