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Here's No. 4 from the 2005 IMO.

Q: Determine all positive integers relatively prime to all the terms of the infinite sequence $$a_n=2^n+3^n+6^n-1, n\geq1$$

I found the solution here, but I'm failing to understand it. It reads:

Trivially, $a_2 = 48$ is divisible by 2 and 3. Okay, I get this, assuming they just picked n=2 since it's a small number.

Let therefore p > 3 be a prime number. Wait what? What does p represent - does it represent n, or n+2? And from the previous step, how can we assume this?

Then $6a_{p−2} = 3·2^{p−1} + 2·3^{p−1} + 6^{p−1} −6 ≡ 3 + 2 + 1−6 ≡ 0 (\mathsf {mod \ p})$. Alone, I think I understand this line, although why multiply by 6? Is $6a_{p-2}$, or any term you would place instead, congruent to $\mathsf {0(mod \ p)}$ since that fulfills the definition of a prime number?

And so the only such number is 1. Only such number refers to what number? Proved. So... what would be the answer?

Thank you for helping!

Eri
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    That link doesn't appear to work. here is another. Perhaps the solution given there will help? Seems clear enough. – lulu May 31 '18 at 17:08
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    The logic of the proof is this: there are no such integers $n>1$. To see that, it suffices to show that no prime $p$ works, since any integer $>1$ is divisible by some prime. The cases $p=2,3$ are handled the way you say. The case of $p>3$ is handled by playing with Fermat's Little Theorem. – lulu May 31 '18 at 17:11
  • @lulu Thank you, that solution is a bit clearer. Here's the link to the solution I was looking at before: https://klevas.mif.vu.lt/~alkauskas/MP3/2005%20solutions.pdf – Eri May 31 '18 at 17:13
  • @lulu Thank you for the logic, it's making a bit more sense to me. I'm confused on what you mean when you say that 'there are no integers $n \geq 1$' – Eri May 31 '18 at 17:18
  • I said "there are no such integers $n>1$" . That "such" is important. I meant "the claim, which requires proof, is that there are no integers $n>1$ which are relatively prime to all of the $a_k$." – lulu May 31 '18 at 17:19
  • That solution is essentially the same as the one in my link, but it is considerably more terse. Study the one I sent. – lulu May 31 '18 at 17:21
  • @lulu I think I've got the whole thing now. I'll start paying attention to small words like 'such' too. Thank you so much! – Eri May 31 '18 at 17:39

1 Answers1

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If p>3 then, $2^{p-2}+3^{p-2}+6^{p-2}\equiv 1\mod p$. Now, multiply both the sides by 6 to get: $$3 · 2^{p−1} + 2 · 3^{p−1} + 6^{p−1} ≡ 6 \mod p$$

this is clearly a consequence of Fermat’s little theorem. Therefore p divides $a_{p−2}$. Also, 2|$a_1$ and 3|$a_2$. So, there is no number other than 1 that is relatively prime to all the terms in the sequence.

Second approach similar to the above one. I think you have doubt with this point.

As you said, $a_2 = 48$ and it is divisible by 2 and 3. I hope you got this point.Then let us take some p > 3 be a prime number.(P is just an assumption, so don't get confused by this).

Then, $$6a_{p−2} = 3 · 2^{p−1} + 2 · 3^{p−1} + 6^{p−1} − 6 ≡ 3 + 2 + 1 − 6 ≡ 0\mod p$$

And so the only such number is 1.

The "Such Number" Here refers to the number which is relatively prime to all terms in the infinite sequence.

Identicon
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