2
$\begingroup$

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.

  • 4
    use that $$\frac{(2^n)^3+n}{2^n+n}=(2^n)^2-n2^n+n^2-\frac{n^3}{2^n+n}$$ – 2017-01-08
  • 3
    And $\displaystyle \frac{10^3}{2^{10}+10}<1$. – 2017-01-08
  • 0
    Beware that the equation in @Dr.SonnhardGraubner's comment is incorrect. See my answer for one correct way. Using modular arithmetic is simpler so helps to avoid such computational errors (since we need only the remainder we can ignore the crufty quotients). – 2017-01-09
  • 0
    All we need in @Dr.SonnhardGraubner's equation is to change the numerator in the remainder term to $n^3-n$. We still are forced to accept $n<10$, else the renainder term is strictly between $0$ and $1$. – 2017-01-09

2 Answers 2

3

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

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

1

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 $a \ge 4$.