I was asked to find all the prime numbers in these forms and prove that these are the only prime numbers in these forms. From some basic research on the internet, I suspect it may have been a trick question and that it is conjectured that they are infinite primes in this form.

I am aware there is a unique prime in the form $n^2-1$, $n^3-1$, using the $(n-1)$ factorisation.

Please would you help.

EDIT: this is meant to be a generalisation for any positive integers, n and m.


closed as unclear what you're asking by TMM, Frits Veerman, jgon, Namaste, Shailesh Oct 5 '16 at 0:39

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    $\begingroup$ This question is a bit odd, to ask especially when we do not even know the finiteness of Fermat primes $2^{2^n} + 1$. $\endgroup$ – астон вілла олоф мэллбэрг Oct 4 '16 at 11:28
  • $\begingroup$ @астонвіллаолофмэллбэрг Or Mersenne primes, or $n^2+1$ primes. $\endgroup$ – Parcly Taxel Oct 4 '16 at 11:29
  • $\begingroup$ Exactly. I added the Fermat primes because they seem very rare compared to the other sequences. $\endgroup$ – астон вілла олоф мэллбэрг Oct 4 '16 at 11:30
  • $\begingroup$ Maybe the question is to find all factors of $n^m \pm 1$, similar to how $n^2 - 1$ factors as $(n-1)(n+1)$? (I'm just guessing though - otherwise the question just does not make sense.) $\endgroup$ – TMM Oct 4 '16 at 12:20

there is an infinite number of primes of the form $n^1 + 1$

if there was a finite set in the form $n^1 + 1$, then multiplying them all together and adding $1$ would also be of the form $n^1 + 1$ for some new n, and it would be a new prime number, not in the original set - so it could never be true

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    $\begingroup$ The intent is certainly that $m$ is an integer greater than $1$. That OP does not make it explicit is maybe worth a comment, yet not giving an answer based on a loophole. $\endgroup$ – quid Oct 4 '16 at 12:06
  • $\begingroup$ @quid - well it's true that I didn't provide an answer to a question if additional info is added by someone else later - the edit to the question seemed to clarify your point though $\endgroup$ – Cato Dec 29 '16 at 11:24

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