Find all $n \in \mathbb{N}$ for which $(2^n + n) | (8^n + n)$.

Question

Find all $n \in \mathbb{N}$ for which $(2^n + n) | (8^n + n)$.

$n = 1, 2, 6$ are some solutions. Also, if the above holds then

$$(2^n + n) | 2^n(2^n-1)(2^n+1)$$

and

$$(2^n + n)| n(2^n+1)(2^n-1)$$

I've tried using cases when $n$ is even or odd and have tried using modular arithmetic but am not able to proceed. Please help.

Thanks.

Answer 1

Hint $\,\ 2^{\large n}\!+n\mid n+\color{#0a0}{8^{\large n}}\!\! \iff 2^{\large n}\!+n\mid n\color{#0a0}{-n^3}\ $ since

$\ {\rm mod}\,\ 2^{\large n}\!+n\!:\,\ \color{#c00}{2^{\large n}\equiv -n}\,\Rightarrow\, \color{#0a0}{8^{\large n}}\!= 2^{\large 3n}\!= (\color{#c00}{2^{\large n}})^{\large 3}\!\equiv (\color{#c00}{-n})^{\large 3}\!\equiv\color{#0a0}{-n^3}$

Answer 2

More generally, if $(a^n+n) |((a^3)^n+n)$, then, since $(a^3)^n+n^3 =(a^n)^3+n^3 =(a^n+n)((a^2)^n-na^n+n^2) $, we have $(a^3)^n+n =(a^3)^n+n^3-n^3+n =(a^n+n)((a^2)^n-na^n+n^2)-n^3+n $ so $(a^n+n) | (n^3-n) $.

Therefore $a^n+n \le n^3-n$, which bounds $n$ as a function of $a$.

In particular, $a^n < n^3$, so $a < n^{3/n}$.

Since $n^{3/n} < e^{3/e} < 3.1 $ for all $n$, $4^{3/4} < 3$, and $n^{3/n} < 2$ for $n \ge 10$, we get these bounds on $n$:

If $a = 2$, $n < 10$. Trying $\dfrac{8^n+n}{2^n+n}$, the only integer values are, according to Wolfy, $(1, 3), (2, 11), (4, 205)$, and $(6, 3745)$.

If $a = 3$, $n \le 3$. Trying $\dfrac{27^n+n}{3^n+n}$ we get, again according to Wolfy, the only solution is $(1, 7)$.

There are no solutions for $a \ge 4$.