Famous Catalan conjecture implies that difference of power is 1 for only trivial solutions. One can generalize this and ask

For given constant $c$, is a difference of two perfect power $c$ for only finitely often?

This is likely to be true and has a name followed by Generalized Catalan conjecture.

Difference of perfect power had several more interesting questions, such as Fermat-Catalan conjecture and many other Diophantine equations.

Given this point of view, one can ask more about the distribution of the difference of two perfect powers.

Now here is my question.

Is difference of two perfect power a prime infinitely often?

Notice that difference of two perfect power can be $4k+1$ because $(2k+1)^2-2k^2=4k+1$. Since there are infinitely many primes in form of $4k+1$, It is trivial in this sense. So I would instead ask,

Is difference of two perfect powers (that are not squares) a prime infinitely often?

I also notice that $k^3-1$ is squarefree infinitely often, so that will answer the case if one asks the easier variation of this question and replace prime with squarefree.

Are there any tries or results with this sortof topics? Thanks.

  • $\begingroup$ Are the exponents of the perfect powers arbitary ? $\endgroup$ – Peter Apr 3 '18 at 8:56
  • $\begingroup$ The Bunyakovsky-conjecture for example implies that there are infinite many positive integers $n$, such that $n^5-3^3=n^5-27$ is prime. $\endgroup$ – Peter Apr 3 '18 at 9:00
  • $\begingroup$ Yeah they are arbitrary. They just have to be a perfect power that is not square. $\endgroup$ – Simo Ryu Apr 3 '18 at 9:01
  • $\begingroup$ I also know that there are a lot of stronger version of this conjecture, such as the one you mentioned. But they are far more difficult to answer because I'm asking for all kinds of polynomials possible. $\endgroup$ – Simo Ryu Apr 3 '18 at 9:04
  • $\begingroup$ I do not know whether it can be proven, but I am pretty sure that the answer is yes. $\endgroup$ – Peter Apr 3 '18 at 9:04

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