$\forall{x}\in\mathbb{Z}$, if $\forall{x}\in\mathbb{N}$, $a^x\mid{x}$, then $a=1$ or $a=-1$.

Question

Prove or disprove:

$\forall{x}\in\mathbb{Z}$, if $\forall{x}\in\mathbb{N}$, $a^x\mid{x}$, then $a=1$ or $a=-1$.

I tried proving the positive but I'm not sure if it is the right approach.

Answer 1

We have: if $a \ge 2 \implies a^x \ge 2^x > x\implies a^x > x\implies a^x \nmid x$. Thus if $a^x \mid x \implies a = 1$. Also, if $a \le -2 \implies |a^x| \ge 2^x > x \implies |a^x| > x\implies |a^x| \nmid x\implies a^x \nmid x$. Thus $a^x \mid x \implies a = -1$. Together we have: $a = \pm 1$.

Answer 2

Note that if $a^x | x,$ then $ a^x \leq x.$ Now consider when this is possible and try to take it from there.

Answer 3

If $a > 1$ is should be become apparent that $a^m > m$. After all $a^m = a*a*a.... $ and as $a*a = a+a+a+a... \ge a+a$ we have $a*a*a* \ge a+a+a+.... > 1+1+1+.... = m$.

(To make this clearer consider $5^3 = 5*5^2 = 5^2 + 5^2 + 5^2 + 5^2 +5^2 \ge 5^2 + 5^2 > 1 + 5^2 = 1 + 5*5 = 1 + 5+5 +5 +5+5 \ge 1+ 5+5 > 1+1+1 = 3$)

We can try to prove that more formally.

If $a > 1$ then $a \ge 2$ and $a^m \ge 2^m =(1+1)^m =_{\text{by binomial theorem}} \sum_{k=0}^m {m\choose k}1^k \ge \sum_{k=0}^m \ge m+1$.

Now for positive numbers if $a|b$ then $a\le b$ so $a^m|m$ means $a^m \le m$ but that's a contradiction if $a > 1$.

So if $a$ is a positive integer and $a^m|m$ then $a =1$.

But what about $0$ and negative values?

Well $0 \not \mid k$ if $k\ne 0$ so that $a \ne 0$.

And for negative numbers $a|b \iff \small{|}a\small{|}\large{\mid}\small{|}b\small{|}$ and if $a < 0$ then $a^m = \pm |a|^m$. So this holds for negative $a$ as well.