How many solutions exist for ${a^p} - a = {n^p}$?

Question

$a,n,p \in \mathbb{Z} , p \ge 0.$ Values cannot be assumed. I have no clue where to even start.

Answer 1

Trivial pairs are $a=n=0$.

Assume $a>0, n>0$, obviously, if $a^p-a=n^p$, then $a>n$.

• $p=0, 1-a=1 \Longrightarrow a=0$, but $0^0$ doesn't make sense.
• $p=1, a-a=n \Longrightarrow n=0$.
• $p \ge 2, (n+1)^p-n^p= \binom{p}{1}n^{p-1}+ \cdots + \binom{p}{p}n^0 \gt n+1$, note that here $n \ge 1$,then $a^p-a \ge (n+1)^p-(n+1) \gt n^p$.

So, no more solution for $a>0, n>0$.

Then cosider $a,n \in \mathbb Z - \{0\}$, here $|a| \ge 1, |n| \ge 1$.

$|a^p-a| \ge |a^p| - |a| \ge |a|^p - |a|$. Moreover, $|a|^p-|a| \le |a^p-a| \le |a|^p+|a|$ and $|n^p|=|n|^p$.

• $|a| > |n|$, similar with we talked above, $|a^p-a| \gt |n^p|$.
• $|a| = |n|$, obviously $|a^p-a| \ne |n^p|$.
• $|a| < |n|$, then $|a^p-a| \le |a|^p+|a| < |a|^p+|n| \le |n-1|^p + |n| \lt (|n|^p - |n|) + |n| = |n^p|$.

Then we infer that $|a^p-a| \ne |n^p|$. We have solved it on $\mathbb Z-\{0\}$.

As for $a=0$ or $n=0$. Firstly take $a=0$, then $p$ should exclude $0$, $n=0$. Secondly, $n=0 \Longrightarrow a^p-a=0 \Longrightarrow a(a^{p-1}-1)=0$. We have

$$\begin{eqnarray} \begin{cases} a = \pm 1,&p \in \{z|z=2k-1, k\in \mathbb Z^+\} \cr a = 1, & p \in \{z|z=2k, k \in \mathbb Z^+\} \cr \end{cases} \end{eqnarray}$$

Conclude, $a=n=0$ for $p \in \mathbb Z^+$. $n=0, a \in \mathbb Z$ for $p=1$. $a=\pm1,n=0$ for $p \in \{z|z=2k-1, k\in \mathbb Z^+\}$. $a=1,n=0$ for $p \in \{z|z=2k, k \in \mathbb Z^+\}$.

Answer 2

A hinted idea:

Work modulo $\;p\;$ (a prime) , so that any integer solution would have to fulfill

$$a^p-a=n^p\iff(a-n)^p=a\iff a-n=a$$