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 $n$ .

#NumberTheory #Sharky

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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.

Department 8 - 5 years, 7 months ago

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

Sharky Kesa - 5 years, 7 months ago

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

Department 8 - 5 years, 7 months ago

@Department 8 – Sorry once more. Edited.

Sharky Kesa - 5 years, 7 months ago

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

Department 8 - 5 years, 7 months ago

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

A Former Brilliant Member - 5 years, 7 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,

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

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

Steven Yuan - 5 years, 7 months ago

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

Department 8 - 5 years, 7 months ago

IMO 2005? O.o

Manuel Kahayon - 4 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$