Infinitely many solutions to an equation with primes: $pqr+22=s$? (open problem)

Solution 1:

THIS IS NOT AN ANSWER, it is just a comment maybe not impertinent and it is put here for obvious reasons of space.

It is clear that $2$ must be discarded. For all pair of odd primes $p,q$ the diophantine equation $px-qy=22$ has infinitely many integer solutions in such a way that there are infinitely many primes for $x$ and $y$ (Dirichlet). The point is that for our purpose we need these primes appear simultaneously (which could deserve to be an interesting problem maybe). We look at some examples.

$$3x-5y=22\iff(x,y)=(5n+4,3n-2)\Rightarrow n=3 \space\text {gives }(r,s)=(19,7)\\5x-7y=22\iff(x,y)=(7n+3,5n-1)\Rightarrow n=4 \space\text {gives }(r,s)=(31,19)\\13x-17y=22\iff(x,y)=(17n+3,13n+1)\Rightarrow n=4 \space\text {gives }(r,s)=(19,7)$$ It must be said that it could be the case that only one of this class of equations is sufficient (to show which would ultimately be a refinement of Dirichlet's Theorem). Another thing is that we have not taken into account the required eighth and ninth position but have limited ourselves to finding four odd primes.

Solution 2:

Comment:

Based on finding of OP by computer using prime generator $15m+22$ and my comment that all primes are of the form $30k+r$ where $r=1, 7, 11, 13, 17. 19, 23, 29$ we solve following Diophantine equation for every r:

$30k+r=15m+22$$\space\space\space\space\space\space\space (1)$

Here for example I solve it for $r=7$:

$p=30k+7=15m+22$

Or:

$2k-m=1$

General form of solutions are:

$(k, m)=(t+2, 2t+3)$

Such that:

$p=30(t+2)+7=15(2t+3)+22=30t+67$

In this way prime generator $p=30t+67$ can give primes satisfying the condition, some examples:

$t=1\rightarrow p=107=22+5\times17$

$t=2\rightarrow p=127=22+5\times 19$

We may try other values for r and solve equation (1) such that the result is $p=22+pqr$ and show experimentally that there can be infinite such primes.