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

For each positive integer nn, let an=2n+3n+6n1a_n = 2^n + 3^n + 6^n - 1. Find all positive integers mm where m<anm<a_n such that gcd(m,an)=1gcd(m,a_n)=1 for all positive integers nn.

Note by Sharky Kesa
3 years, 11 months ago

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For every positive integer nn there will be I think there would infinitely many mm as:-

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

We see that for all nn the sequence Input will be even and there will be infinitely many mm 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 nn because there is no specific pattern for ana_{n}, the Euler's Totient Function of ana_{n} will keep changing, by using wolfram alpha.

Department 8 - 3 years, 11 months ago

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

Sharky Kesa - 3 years, 11 months ago

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I think this will also give infinity until you mention mm is a positive integer less than nn

Department 8 - 3 years, 11 months ago

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@Department 8 Sorry once more. Edited.

Sharky Kesa - 3 years, 11 months ago

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

Department 8 - 3 years, 11 months ago

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

Anmol Agarwal - 3 years, 11 months ago

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

First, 2,348=a22, 3|48 = a_2. Now, let's assume p>3p > 3. Then,

6ap26(2p2+3p2+6p21)3(2p1)+2(3p1)+6p163+2+160(modp). \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, p6ap2p|6a_{p - 2}. Since pp is not 2 or 3, we must have pap2p|a_{p - 2}, and our proof is complete.

Steven Yuan - 3 years, 11 months ago

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

Department 8 - 3 years, 11 months ago

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IMO 2005? O.o

Manuel Kahayon - 2 years, 11 months ago

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