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Determine all positive integers relatively prime to all the terms of the infinite sequence $a_n = 2^n + 3^n + 6^n − 1,\ n \ge 1$. ~IMO 2005 P4

I was solving a number theory book in which this was an example question. I know I can find this solution anywhere but I would like a hint as to how to solve this problem coz i don't want to see the solution just yet. I think this may use Fermat's little theorem so I was looking at $a_{p-1}$ . Any hint would be appreciated! Cheers.

Bill Dubuque
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Aditya_math
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1 Answers1

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Modulo a prime $p\ge5$, $a\cdot2^{p-1}+b\cdot3^{p-1}+c\cdot6^{p-1}\equiv a+b+c$ by Fermat's Little Theorem. Choose $a,\,b,\,c$ so you can divide this by $6$ to prove $p$ is a factor of some $a_n$. What about $p\in\{2,\,3\}$?

J.G.
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  • wow this is cool, so i choose a,b,c as 3,2,1 and then divide by 6 to get $a_{p-2}$ is a factor of p for p>=5. Thanks! – Aditya_math Nov 16 '20 at 08:07
  • i just have one more question, how would one stumble upon this. i am assuming you have solved this problem, so how did you originally think of this? – Aditya_math Nov 16 '20 at 08:11
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    @Aditya_math Working out which integers a sequence is coprime to boils down to finding which primes, if any, can't divide it. So if you want to eliminate candidates with a sequence like this, FLT is a natural starting point in the IMO toolkit. But $a_{p-1}$ isn't an obvious multiple, so you have to try a nearby sequence element. – J.G. Nov 16 '20 at 08:16
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    @Aditya_math If I was attacking this problem, it might have taken me 12 hours, on and off to solve. My first step would have been to write a computer program to manually compute $a^n \pmod{b}$ where $a \in {2,3,6}$ and $b \in {2,3,5,7,11,13,17,19}.$ Ultimately, I would focus on a math rather than computer solution, but first, I would look for patterns in the data. Then, I might google something like: number theory congruence theorem. – user2661923 Nov 16 '20 at 08:49
  • @user2661923 unfortunately we can't use computers in math contests lol – Aditya_math Nov 16 '20 at 08:52
  • @Aditya_math Nor can you spend 12 hours on the problem. Which is (for some of us) a good reason to avoid direct participation in contests, but still examine the contest problems at home, in our leisure. Better to do math to live than live to do math. – user2661923 Nov 16 '20 at 08:55
  • @user2661923 computer science 101 – Aditya_math Nov 16 '20 at 08:56