Let $(a,b)$ be a pair of coprime positive integers with $a$ being even. Are these conditions sufficient to prove that there exist infinitely many positive integers $n,$ such that $(a^n+1,b^n+1)=1$ ?
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2$\begingroup$ "Sufficient to prove that" is an interesting formulation :) If it is taken literally, then I am clueless about the answer since nobody has been able to prove this. In other words, this is a well known conjecture. However, using the work of Corvaja and Zannier based on Schmidt's Subspace theorem, one can at least prove the sub-exponential upper bound $\exp(\varepsilon n)$ on this gcd, for all $\varepsilon > 0$, and infinitely many $n$. (Indeed all large enough $n$ if additionally $a$ and $b$ are multiplicatively independent.) $\endgroup$– Vesselin DimitrovCommented Jul 31, 2019 at 8:27
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$\begingroup$ If you can prove there is one n, you should be able to prove there are infinitely many. Gerhard "Start With A Simpler Case" Paseman, 2019.07.31. $\endgroup$– Gerhard PasemanCommented Jul 31, 2019 at 8:33
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$\begingroup$ Is it at least true that there are infinitely many values of $n$ with $(2^n+1,3^n+1)=1$? $\endgroup$– SevaCommented Jul 31, 2019 at 8:51
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$\begingroup$ Thank you very much for enlightening me. I was almost sure that it is a famous conjecture, but I couldn't find any papers about it. $\endgroup$– Kristiyan VasilevCommented Jul 31, 2019 at 9:17
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$\begingroup$ Is it known that for any set $P$ of primes of positive relative density ($|P\cap[1,x]\gg x/\log x$) there are infinitely many exponents $n$ such that $b^n+1$ is divisible only by the primes from $P$? Choosing $P$ to be the set of all primes $p\equiv 3\pmod 4$ such that $a$ is a quadratic residue modulo $p$, for every such $n$ we will have $(a^n+1,b^n+1)=1$ since $a^n+1$ is not divisible by primes $p\in P$ in view of $a^n\equiv -1\pmod p$. $\endgroup$– SevaCommented Jul 31, 2019 at 12:41
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I think that this sort of question was originally asked by Ailon and Rudnick, but they use $-1$ instead of $+1$ and asked if $\gcd(2^n-1,3^n-1)=1$ for infinitely many $n$. In this setting, for more general $a$ and $b$, the right question/conjecture would be $\gcd(a^n-1,b^n-1)=\gcd(a-1,b-1)$. They prove something stronger if you replace $\mathbb Z$ with $\mathbb C[t]$. Here's the Ailon-Rudnick paper:
N. Ailon and Z. Rudnick, ‘Torsion points on curves and common divisors of $a^k-1$ and $b^k-1$, Acta Arith., 113 (2004), no. 1, 31–38 (MSN).
There have been lots of articles on this and related problems. For example, a Google search on "Ailon Rudnick gcd" brings up recent articles such as:
On some extensions of the Ailon–Rudnick theorem, A Ostafe - Monatshefte für Mathematik, 2016 (MSN)
On a variant of the Ailon–Rudnick theorem in finite characteristic. D Ghioca, LC Hsia, TJ Tucker - New York Journal of Mathematics, 2017 (MSN)
Greatest common divisors of iterates of polynomials, LC Hsia, T Tucker - Algebra & Number Theory, 2017 (MSN)
answered Jul 31, 2019 at 11:47