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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Jendrik Stelzner
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Kai
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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. – LPenguin Aug 28 '18 at 04:01
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Thanks LPenguin. $a^p+p$ divides $b^p+p$, p prime, $gcd(a,p)=gcd(b,p)=1$ – Kai Aug 28 '18 at 04:44
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That means $k(a^p+p)=b^p+p$ – Kai Aug 28 '18 at 04:45
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For k is a natural number, k>1. (I am trying to prove by contradiction) – Kai Aug 28 '18 at 04:46
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Therefore $ka^{p}+kp=b^p+p$ – Kai Aug 28 '18 at 04:52
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$(k-1)(a^p+p)=b^p-a^p$ – Kai Aug 28 '18 at 04:53
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This is congruent to b-a mod p – Kai Aug 28 '18 at 04:55
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By Fermat Little Theorem – Kai Aug 28 '18 at 04:56
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$(k-1)a$ is congruent to $b-a$ mod p – Kai Aug 28 '18 at 04:56
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$ak$ is congruent to $b$ mod p – Kai Aug 28 '18 at 04:58
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What should I do? – Kai Aug 28 '18 at 04:59
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This question is a near duplicate, and has a good answer. – Fimpellizzeri Aug 28 '18 at 13:24
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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 $$a^n \equiv a\equiv -k(p-1)\pmod{p} \implies p\mid a^n+n \mid b^n+n \\ \implies a^n \equiv b^n\pmod{p} \implies a \equiv b\pmod{p} \implies a=b$$
tarit goswami
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1I 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. – Kai Aug 28 '18 at 17:51
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@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! – tarit goswami Aug 28 '18 at 17:55
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@Kai You have posted some comments about your approach, it will be better to write then in the problem. – tarit goswami Aug 28 '18 at 18:25