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

2 Answers 2

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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 $p \in \{z|z=2k, k \in \mathbb Z^+\}$.

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    we work in $\mathbb Z$ – 2017-01-12
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    @zwim I know it. But I haven't come up with a solution for $\mathbb Z$. – 2017-01-12
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    This general idea is IMHO the way to do it. I think that you left out the case that $p$ is even and $a$ is negative, when $a^p-a>a^p$. But, analogously to your strategy you can check that (at least most of the time) $n=|a|+1$ is already too large. – 2017-01-12
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    So $\Large{+1}$ – 2017-01-12
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    @JyrkiLahtonen Yeah, thanks for your correction. – 2017-01-12
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    Isn't there any issue ? I agree $a^p\ge(n+1)^p$ and $a\ge(n+1)$ but not with $a^p-a\ge(n+1)^p-(n+1)$. – 2017-01-12
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    Being a 9th grade student, the inequalities completely flew over my head. Can you clarify? – 2017-01-12
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    @zwim $x^p-x$ is increasing when $x \ge 1$, you can easily verify it. That means $(n+1)^p-(n+1) \gt n^p \gt n^p-n$. – 2017-01-12
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    @MainakRoy The most important step is that $(n+1)^p - n^p \gt n+1$. Here I use a theorem called binomial theorem. You can imagine $(n+1)^p$ is $(n+1) \cdot (n+1) \cdots (n+1)$ which has $p$ (n+1) multiplying with each other. Actually, it increases very fast like $a^p \ll b^p$ when $a \lt b$. With lemma $(n+1)^p-(n+1) \gt n^p$, you can easily prove the several answers. – 2017-01-12
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    I want to give +1 to this solution but you are skipping details. For instance the case $|a|>|n|$ is true ony for $|n|\ge 1$ so you miss solutions $a=1,n=0$ and $a=-1,n=0,p\;odd$. The obviousity of case $|a|=|n|$ is not immediate, and $|a|^p+|n|<|n^p|$ comes abruptly (i.e. $|a|^p+|n|\le |n-1|^p+|n|<(|n|^p-|n|)+|n|=|n^p|$). – 2017-01-12
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    @zwim Yeah, thanks, I'm too careless. – 2017-01-12
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A hinted idea:

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

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

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    How did $a^p - n^p = a$ transform into $(a - n)^p = a$ ? – 2017-01-12
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    @MainakRoy This is basic number theory: in modulo $\;p\;$ a prime, we always have for natural numbers that $\;(w+z)^p=w^p+z^p\pmod p\;$ – 2017-01-12
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    @MainakRoy Go check Fermat's little theorem https://en.wikipedia.org/wiki/Fermat's_little_theorem (damn I forgot how to do links). – 2017-01-12
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    By Fermat's Little theorem $p\mid a^p-a$, so $p\mid n^p$, so by Euclid's Lemma $p\mid n$. – 2017-01-12
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    But it's nowhere mentioned in my question that $p$ is a prime. Then does this imply that the polynomial is unsolvable for a non-prime $p$? – 2017-01-12
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    @MainakRoy If $\;p\;$ is not a prime then I don't know what to do off the top of my head...That's why I wrote explicitly for $\;p\;$ prime. – 2017-01-12