Prove that for infinitely many integers $n>2$ equation
$a^n-(a-2)^n=b^{n-1}$
has no integer solutions for $a,b$.
Edit: I would appreciate any hints. They may concern other nonlinear diophantine equations for n>... as I think I wouldn't be able to do even another example and I guess I may not be the only one having these difficulties
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$\begingroup$ Can someone explain why it got downvoted? $\endgroup$– user509482Dec 3, 2017 at 10:38
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$\begingroup$ Why do you downvote? $\endgroup$– user509482Dec 3, 2017 at 10:47
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1$\begingroup$ A downvote means (official definition on this site) "This question does not show any research effort; it is unclear or not useful." Where's your research? What's your question? Imperatives like "prove", "show", "explain" aren't popular, here. $\endgroup$– user436658Dec 3, 2017 at 11:01
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$\begingroup$ @Professor Vector I got a Diophantine exercise list with equations of exponents 2,3,4 and 5. Then there is this one, that I have no idea how to solve $\endgroup$– user509482Dec 3, 2017 at 11:09
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$\begingroup$ @user509482 Just a curiosity. What kind of school do you attend? I ask because I am a math graduate here in Italy and I never had such assignments... $\endgroup$– RaffaeleDec 3, 2017 at 11:12
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1 Answer
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The equation doesn't have integer solutions, if $n$ is an odd prime: because of the theorem of Fermat, we have $x^{n-1}=1\pmod n$, for $x\neq0\pmod n$, so that always $x^n=x\pmod n$. That means $a^n-(a-2)^n=a-(a-2)=2\pmod n$. But $b^{n-1}\pmod n$ is $0$ if $n$ divides $b$, and $1$ otherwise, so the equation can't be satisfied.
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$\begingroup$ b^(n-1)(mod n) is 0 when b (mod n)=1 and 1 when b (mod n) doesn't equal 1? $\endgroup$ Dec 3, 2017 at 12:51
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$\begingroup$ Sorry, that was a typo, I've edited it. $\endgroup$– user436658Dec 3, 2017 at 12:57
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$\begingroup$ b^(n-1)(mod n) is 0 when b (mod n)=0 and 1 when b (mod n) doesn't equal, am I right? $\endgroup$ Dec 3, 2017 at 13:00