# 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 $$n$$.

Note by Sharky Kesa
3 years, 5 months ago

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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.

- 3 years, 5 months ago

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

- 3 years, 5 months ago

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

- 3 years, 5 months ago

Sorry once more. Edited.

- 3 years, 5 months ago

Can you help me find the no. of factors of $$(2^{n}+1)(3^{n}+1)$$

- 3 years, 5 months ago

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

- 3 years, 5 months ago

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{align*} 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{align*}

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

- 3 years, 5 months ago

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

- 3 years, 5 months ago

IMO 2005? O.o

- 2 years, 5 months ago