Let $f(a,n) = \dfrac{a^{n \frac{n-2}{2}} - 1}{(n-1)(n+1)}$. Then $\Bbb{Z} \times(\Bbb{Z}\setminus\{\pm{1}\}) \xrightarrow{f} \Bbb{R}$ takes on infinitely many integer values for $a \neq 0, \pm 1$ when $n \gt 2$.
Possible to prove?
$\frac{a^{n \tiny\frac{n-2}{2}} - 1}{(n-1)(n+1)} \in \Bbb{Z}$ infinitely often.
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sequences-and-series
elementary-number-theory
divisibility
real-numbers
integers
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0This isn't what you're looking for, but for $n = 2$, $f(a,2) = 0$ for all $a \neq 0$. – 2017-01-06
3 Answers
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How about $f(4n^2,2n)$ for $n>1$? We then have $$ f(4n^2,2n)=\frac{(4n^2)^{2n(n-1)}-1}{(2n-1)(2n+1)}=\frac{(4n^2-1)\sum_{k=0}^{2n(n-1)-1}(4n^2)^k}{(4n^2-1)}=\sum_{k=0}^{2n(n-1)-1}(4n^2)^k\in\mathbb{N}. $$
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For $n = 4$, you are looking at
$$\frac{a^{4}-1}{15}$$
and the question is when $a^4 -1$ is divisible by both $3$ and $5$. You can check that $a^4 -1$ is divisible by $3$ if and only if $a$ is not, and the same goes for $5$. So $f(a,4)$ is an integer if and only if $a$ is not divisible by $3$ or $5$.
A more interesting question might be to ask for which $n$ is $f(a,n)$ never an integer.
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If $n-1$ and $n+1$ are twin primes, and $a$ is not a multiple of none of them, $$a^n\equiv1\pmod {n+1}$$ $$a^{n-2}\equiv1\pmod{n-1}$$
Since $n(n+2)/2$ is a multiple of $n$ and $n+2$, then $$a^{n\frac{n+2}2}\equiv 1\pmod{n^2-1}$$
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0Perhaps this can be used to show that there are infinitely many twin primes! :) – 2017-01-06