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Procrastinal problem: Given $n$, one can ask for integers $a,b>1$ of different parities such that $a+b^k$ is prime for $k=(a\bmod 2),\ldots,n$.

A few examples are:

$2+4995825^k$ is prime for $k=0,\ldots,6$.

$1708+6301^k$ is prime for $k=0,\ldots,8$.

$4503+4^k$ is prime for $k=1,\ldots,14$.

Given $n$, do there exist sets of primes of the form $\lbrace a+b^k,k=(a \bmod 2)\ldots,n\rbrace$?

Do there exist such sets with fixed $a\geq 2$, with fixed $b\geq 2$? Given $n$, can one say something on 'smallest' such sets (say with $a+b$ minimal, or with $ab$ minimal)?

Are such sets necessarily finite? (The existence of such an infinite set of primes would be very surprising, see also Density of primes in sequences of the form $a^n+b$)

(Stupid observation: $x^3+1=(x-1)(x^2-x+1)$ implies that $a=1$ is not interesting.)

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    $\begingroup$ There is no prime $p$ such that $4^k$, $k=1,2,\dots,n$ covers all the residue classes mod $p$ (since $4$ is a quadratic residue mod $p$, its powers can only be quadratic residues), so those powers of four constitute an admissible set. Then standard (but unproved) conjectures imply that for each $n$ there exist $a$ such that $a+4^k$ is prime for $1\le k\le n$. $\endgroup$
    – Gerry Myerson
    Commented Sep 8, 2022 at 5:17
  • $\begingroup$ Thanks Gerry. I do not understand the link between admissibility and primality: Is admissibility a necessary prerequisites for the conjectures? $\endgroup$
    – Roland Bacher
    Commented Sep 8, 2022 at 7:43
  • $\begingroup$ If you have a prime $p$ and a set $\{a_1,a_2,\dots,a_n\}$ of positive integers which covers all congruence classes modulo $p$, then you can't have $m\ge p$ such that $m+a_i$, $1\le i\le n$, are all prime. E.g., $m+2$, $m+4$, $m+6$ can't all be prime (for $m>1$) since one of them will be a multiple of three, but $m+2$, $m+4$, $m+8$ can all be prime and, conjecturally, are all prime for infinitely many $m$. We say $\{2,4,8\}$ is admissible, $\{2,4,6\}$ isn't. en.wikipedia.org/wiki/Prime_k-tuple#Admissibility $\endgroup$
    – Gerry Myerson
    Commented Sep 8, 2022 at 9:18
  • $\begingroup$ Thanks. I was indeed aware of this: $b$ (the basis of powers) tends to have many small prime-divisors. Your observation implies that $b$ is a perfect square if it is fixed and $n$ can be arbitrarily large. $\endgroup$
    – Roland Bacher
    Commented Sep 8, 2022 at 13:48
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    $\begingroup$ Procrastination comment. If I fix $a=2$, I find (with a small PARI/GP function) "good" $b$ values 909, 4995825 (your example), 28212939. Searching OEIS gives A245510; they do not have better examples. $\endgroup$
    – Jeppe Stig Nielsen
    Commented Sep 8, 2022 at 14:49

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