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Let $ (a_n) $ be a sequence of numbers such that for all natural numbers $ n $:

$$ a_n = 2^n+3^n+6^n-1 $$

Show that every prime number divides some number in that sequence.

Bill Dubuque
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Dood
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  • I trust you have made some effort to solve this on your own. Please show this in your question, including what you've had difficulty with. Thanks. – John Omielan Jan 23 '19 at 01:53
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    This is IMO 2005, Problem 4. By the way, I can't see any reason as to why it is closed. – TBTD Jan 23 '19 at 15:07
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    Also, is really seems quite weird that some problems are closed while they really look alright, and those which are missing details remain open... It seems it is highly stochastic, and that, it depends on whoever is reviewing. – TBTD Jan 23 '19 at 15:08
  • Duplicate of this old question – Bill Dubuque Oct 31 '24 at 21:24

1 Answers1

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Let $p$ be prime number. Consider $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1$. Then \begin{align*} 6a_{p-2} & \equiv 6(2^{p-2}+3^{p-2}+6^{p-2})-6 \pmod{p}\\ & \equiv 3\cdot 2^{p-1}+2\cdot 3^{p-1}+6^{p-1}-6\pmod{p}\\ & \equiv 3+2+1 -6\pmod{p}\\ & \equiv 0 \pmod{p}. \end{align*}

For $p>3$, we will have $\gcd(6,p)=1$. So we can multiply by $6^{-1}$ on both sides to get $$a_{p-2} \equiv 0 \pmod{p}$$

Now you can deal with $p=2,3$ as special case.

Robert Israel
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Anurag A
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    Intuitively, taking $n=-1$ would give $a_{-1} = \frac12 + \frac13 + \frac16 - 1 = 0$, and the remainders mod $p$ should repeat every $p-1$ steps, so taking $n=p-2$ also works. This isn't quite rigorous, but it's motivation for considering $a_{p-2}$. – Misha Lavrov Jan 23 '19 at 02:02
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    Thanks, I had to look up how to justify that all these powers were congruent to 1 (mod.p). Indeed Fermats little theorem solves this problem fast. Thanks for help and for such quick response. – Dood Jan 23 '19 at 02:07