# Primes from arithmetic and geometric progressions

The five primes, 131, 157, 211, 349, 739, are neither in arithmetic or geometric progression, but are instead the sum of the five corresponding terms of an arithmetic and geometric progression.

Are there arbitrarily long sequences of primes with this property, that is, that they are the sum of the corresponding terms of an arithmetic and a (non-constant) geometric progression?

https://puzzling.stackexchange.com/questions/89739/primes-from-arithmetic-and-geometric-progressions/89740#89740

• There are arbitrarily long APs in primes, take your geometric progression to be constant and you're good. If you require they are not in arithmetic progression, it gets harder. – Wojowu Oct 2 '19 at 12:46
• @Wojowu I do! I shall clarify the issue. – Bernardo Recamán Santos Oct 2 '19 at 12:54
• $1+2=3$, $3+4=7$, $5+8=13$, $7+16=23$, $9+32=41$ is a smaller set of five. – Gerry Myerson Oct 2 '19 at 13:18
• @Wojowu I wonder how much harder it gets. A non-constant geometric progression could be "almost constant", meaning that all elements are about the same size. Then maybe a modification of Green–Tao will be able to show that there are arbitrarily long arithmetic progression such that the sums with that given geometric progression are all prime. It's just my very vague idea, and I barely have any knowledge in this domain. Perhaps we can at TerryTao. – WhatsUp Oct 2 '19 at 14:34
• A sequence of ten primes: $14+9=23,56+27=83,98+81=179,140+243=383,182+729=911,224+2187=2411,266+6561=6827,308+19683=19991,350+59049=59399,392+177147=177539$ – Freddy Barrera Oct 4 '19 at 12:44

More precisely, given any integer-valued polynomials $$P_1,\cdots, P_k$$ in one unknown $$m$$ all with constant term $$0$$, there are infinitely many integers $$x$$, $$m$$ such that $$x + P_1(m), \cdots, x + P_k(m)$$ are simultaneously prime.
From what I understand, this implies (or means) that there exists positive integers $$x$$ and $$m$$ such that all the $$x + P_i(m)$$ are primes.
Now for any fixed $$k$$, simply take $$P_i(m) = m^i + im$$. We will then have: $$x + P_i(m) = (x + im) + m^i,$$ where $$(x + im)_i$$ and $$(m^i)_i$$ are arithmetic and geometric progressions, respectively.