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 \(n\).

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.

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.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestIMO 2005? O.o

Log in to reply

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.

Log in to reply

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

Log in to reply

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

Log in to reply

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.

Log in to reply

Sorry, I forgot to mention that it was for all positive integers \(n\).

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply