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In the ongoing effort of dealing with abstract duplicates. This question is about the lemma:

Lemma Let $k \ge 2$, $p$ prime and $a$ coprime to $p$. Then $$p^2\!\mid a n^k+ bp\iff p\mid n,b.$$

This lemma answers the following kinds of questions:

Sera Gunn
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    The second claim in blue can be seen easily. $x$ must be divisible by $7$, but then, $7^n-x^2$ is divisible by $49$, if $n>1$ holds. – Peter Dec 05 '19 at 21:21
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    The lemma is not difficult either. Since $a$ is not divisble by $p$, $n$ must be dividible by $p$. This forces $n^k$ to be divisble by $p^2$ , hence $bp$ must be divisible by $p^2$, forcing that $b$ is divisible by $p$. – Peter Dec 05 '19 at 21:26
  • Is the question about proving the lemma? – rtybase Dec 05 '19 at 21:42
  • @rtybase Yes. Bill Dubuque realized that this lemma would solve both statements. Once this question has an answer, we can close both of the linked questions as duplicates of this one. – Sera Gunn Dec 05 '19 at 21:46
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    @TrevorGunn I see. Well, Peter, technically, proved it. – rtybase Dec 05 '19 at 21:47
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    @rtybase To clarify, a proof of the Lemma was already posted last month in one of the other linked questions, but some users objected to using that more specific question as an (abstract) dupe target. – Bill Dubuque Dec 06 '19 at 00:55

1 Answers1

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Lemma $\, \ p^2\!\mid a n^k+ bp\iff p\mid n,b,\ $ if $\ \color{#0a0}{k\ge 2}\,\ $ & $\ a\,$ is $\rm\color{#90f}{coprime}$ to ${\rm\color{#c00}{prime}}\ p$ (or ${\rm\color{#c70}{squarefree}}\ p$)

Proof $\ \ (\Leftarrow)\ \ $ Clear by $\,p\mid b,n\,\overset{\times\, p}\Rightarrow\,p^2\mid bp, n^{\color{#0a0}k}_{\phantom{|_{|_.}}}$ and basic divisibility laws.
$\ (\Rightarrow)\ $ $\,p\mid an^k\! + bp\overset{\color{#90f}{(a,p)=1}}\Longrightarrow\! p\mid n^k\color{#c00}{\overset{\rm EL}\Rightarrow}\, p\mid n\,$ $\overset{\color{#0a0}{k\,\ge\, 2}}\Longrightarrow\,p^2\mid n^k\Rightarrow p^2\mid pb\,\Rightarrow\,p\mid b$

using basic divisibility laws, and $\,\color{#c00}{\rm EL}$ = Euclid's Lemma.

Remark $ $ It also holds true for ${\rm\color{#c70}{squarefree}}\:p\,$ because they are precisely those integers satisfying the key middle inference, i.e. $\ p\mid n^k\color{#c00}\Rightarrow\,p\mid n,\,$ for all integers $\,n$.

Bill Dubuque
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