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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!

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

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Siong Thye Goh, user26857, Jendrik Stelzner, Nosrati, Namaste
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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
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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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