I need help with this question.

Let $a$ and $b$ be positive integers such that $a^{n}+n$ divides $b^{n}+n$ for every natural number $n$. Show that $a=b$.

Any help would be appreciated! Thanks!


closed as off-topic by Siong Thye Goh, user26857, Jendrik Stelzner, Nosrati, Namaste Aug 28 '18 at 12:28

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  • $\begingroup$ You could try something along the lines of choosing $n$ such that $n$ is a prime not in the prime factorization of either of $a$ or $b$. Then use the fundamental theorem (arithmetic) to show $a=b$ is a necessary condition. $\endgroup$ – LPenguin Aug 28 '18 at 4:01
  • $\begingroup$ Thanks LPenguin. $a^p+p$ divides $b^p+p$, p prime, $gcd(a,p)=gcd(b,p)=1$ $\endgroup$ – Kai Aug 28 '18 at 4:44
  • $\begingroup$ That means $k(a^p+p)=b^p+p$ $\endgroup$ – Kai Aug 28 '18 at 4:45
  • $\begingroup$ For k is a natural number, k>1. (I am trying to prove by contradiction) $\endgroup$ – Kai Aug 28 '18 at 4:46
  • $\begingroup$ Therefore $ka^{p}+kp=b^p+p$ $\endgroup$ – Kai Aug 28 '18 at 4:52

The main idea came to me when I have solved this problem is that, if we can show that any number which is greater than $b-a$ divides $b-a$, then we will have $b=a$. Let $p$ is a prime satisfies $\gcd(a,p)=1$ , $\gcd(b,p)=1$ and $p>b-a$. Consider $k$, any positive integer such that $p\mid k-a$. As, it is true for any natural number, consider a $n=k(p-1)+p$. Fermat's Little Theorem gives $$a^n \equiv a\equiv -k(p-1)\pmod{p} \implies p\mid a^n+n \mid b^n+n \\ \implies a^n \equiv b^n\pmod{p} \implies a \equiv b\pmod{p} \implies a=b$$

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    $\begingroup$ I figured out another solution. Let p be prime, and a divide p and let b not divide p. This way $a^p+p$ is divisible by p, but $b^p+p$ is not. Contradiction. $\endgroup$ – Kai Aug 28 '18 at 17:51
  • $\begingroup$ I came up with this two hours after I asked. $\endgroup$ – Kai Aug 28 '18 at 17:52
  • $\begingroup$ @Kai Well done, please accept the answer if this one fulfills what u have wanted, otherwise it will be treated as unanswered question and will come back to the top of question queue again and again! $\endgroup$ – tarit goswami Aug 28 '18 at 17:55
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    $\begingroup$ Well I think my answer is right. I got it. It is an answered question. $\endgroup$ – Kai Aug 28 '18 at 18:21
  • $\begingroup$ @Kai You have posted some comments about your approach, it will be better to write then in the problem. $\endgroup$ – tarit goswami Aug 28 '18 at 18:25

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