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Can you prove the following:

Conjecture:  Let $\ p\in\mathbb P\ $ be an arbitrary prime. Then there exist two relatively prime integers $\ a\ $ and $\ b\ $ such that $\ a>0\ $ and $\ b>1\ $ and

$$ \frac{b^p-a^p}{b-a} \in\ \mathbb P. $$

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A harder question would be a restriction to $\ b=a+1.$

See also the comment below by @GHfromMO.

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Finally, one may ask the same question about $\ 2\, <\, p\, \in\,\mathbb P\ $ but for:

$$ \frac {a^p+b^p}{a+b}. $$

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  • $\begingroup$ Would you add some context? Is this a conjecture found somewhere, or a conjecture you formulated? Are you asking whether this is true, or whether this is known (as a conjecture)? $\endgroup$
    – YCor
    Commented Mar 25, 2022 at 8:36
  • $\begingroup$ @YCor, yes, I've formulated it. #### No, I never saw it anywhere at any time; this conjecture is so simple that there is a good chance that I was not the first. $\endgroup$
    – Wlod AA
    Commented Mar 25, 2022 at 8:45
  • $\begingroup$ What is the question exactly? $\endgroup$
    – Piotr Achinger
    Commented Mar 25, 2022 at 8:45
  • $\begingroup$ @PiotrAchinger, "Conjecture" is a synonym of "question" (while "conjecture" is more specific than a general "question"). #### The question is: can you prove it? $\endgroup$
    – Wlod AA
    Commented Mar 25, 2022 at 8:48
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    $\begingroup$ The conjecture is a special case (or weaker version) of Bunyakovsky's conjecture: there should be infinitely many solutions even with $a=1$ (for any given $p$). I am certain that even this special case (or weaker version) is out of reach. See en.wikipedia.org/wiki/… $\endgroup$
    – GH from MO
    Commented Mar 25, 2022 at 9:04

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