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!
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.