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What I did so far: I imagened a polynomial $P(x)$ with degree of $n$. $P(6)=a_n6^n + \dots + a_26^2 + a_16 + a_0$

$P(2)=a_n2^n +\dots+ a_22^2 + a_12 + a_0$

$P(6) - P(2) =a_n(6^n - 2^n) +\dots + a_2(6^2 -2^2) + a_1(6 - 2) +0$

I have found out that $a^n - b^n$ will always be divisible by $a-b$. If I can prove that, then I know that $P(6) - P(2)$ has to be divisible by $6-2=4$. But I don’t know how to easily prove that.

Bill Dubuque
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Roger Powell
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1 Answers1

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You are around the corner for the answer. What you showed is that $4 = (6-2) \mid (P(6) - P(2)) = 6 - 4 = 2$ which is a contradiction. So no such polynomial exist.

Wang YeFei
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  • Yeah I know that its contradiction.I pardon for not being more precise.I am not sure how to prove that $a^n - b^n$ will always be divisible by $a-b$ – Roger Powell Mar 20 '24 at 18:38
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    @RogerPowell setting $a=b$ gives $0$ and thus $a-b$ must be a factor – UnsinkableSam Mar 20 '24 at 18:41
  • Yeah.It was so simple that I am emberessed.Even chatgpt could undersyand it.Thank you a lot. – Roger Powell Mar 20 '24 at 18:50
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    Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Mar 20 '24 at 21:47