Some expression you may have seen before but not in this manner

For each positive integer $n$, let $a_n = 2^n + 3^n + 6^n - 1$. Find all positive integers $m$ where $m<a_n$ such that $gcd(m,a_n)=1$ for all positive integers $n$.

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For every positive integer $n$ there will be I think there would infinitely many $m$ as:-

$\large{a_{n}=(2^{n}+1)(3^{n}+1)-2}$

We see that for all $n$ the sequence Input will be even and there will be infinitely many $m$ which will satisfy it.

Sharky Kesa , If I am wrong somewhere please enlighten me.

Well after the editing of question done by Sharky Kesa , I believe there will be no specific answer for any $n$ because there is no specific pattern for $a_{n}$, the Euler's Totient Function of $a_{n}$ will keep changing, by using wolfram alpha.

I've seen this problem before, so I already know the answer. :\ Nevertheless, I'll post a solution here.

We will prove that $m = 1$ is the only such integer. Let $p$ be a prime number. If we prove that there exists an $n$ such that $p|a_n$, we are done, since that means $m$ can't have any prime factors, so it is equal to 1.

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## Comments

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TopNewestFor every positive integer $n$ there will be I think there would infinitely many $m$ as:-

$\large{a_{n}=(2^{n}+1)(3^{n}+1)-2}$

We see that for all $n$ the sequence Input will be even and there will be infinitely many $m$ which will satisfy it.

Sharky Kesa , If I am wrong somewhere please enlighten me.

Well after the editing of question done by Sharky Kesa , I believe there will be no specific answer for any $n$ because there is no specific pattern for $a_{n}$, the Euler's Totient Function of $a_{n}$ will keep changing, by using wolfram alpha.

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Sorry, I forgot to mention that it was for all positive integers $n$.

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I think this will also give infinity until you mention $m$ is a positive integer less than $n$

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$(2^{n}+1)(3^{n}+1)$

Can you help me find the no. of factors ofLog in to reply

I think for a more general ans. let's try putting values.

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I've seen this problem before, so I already know the answer. :\ Nevertheless, I'll post a solution here.

We will prove that $m = 1$ is the only such integer. Let $p$ be a prime number. If we prove that there exists an $n$ such that $p|a_n$, we are done, since that means $m$ can't have any prime factors, so it is equal to 1.

First, $2, 3|48 = a_2$. Now, let's assume $p > 3$. Then,

$\begin{aligned} 6a_{p - 2} &\equiv 6(2^{p - 2} + 3^{p - 2} + 6^{p - 2} - 1) \\ &\equiv 3(2^{p - 1}) + 2(3^{p - 1}) + 6^{p - 1} - 6 \\ &\equiv 3 + 2 + 1 - 6 \\ &\equiv 0 \pmod{p}. \end{aligned}$

So, $p|6a_{p - 2}$. Since $p$ is not 2 or 3, we must have $p|a_{p - 2}$, and our proof is complete.

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Please tell me whether I am right or not I used Wolfram Alpha to check this out.

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IMO 2005? O.o

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