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selfluminous
 Post subject: arithmetic problem
PostPosted: Sat, 28 Jan 2012 05:27:21 UTC 
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prove that if 1+2^n+4^n is a prime number, then n is a power of 3

im not sure where to start with this kind of problem
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outermeasure
 Post subject: Re: arithmetic problem
PostPosted: Sat, 28 Jan 2012 08:21:09 UTC 
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Topic moved from Calculus to A&NT.

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Shadow
 Post subject: Re: arithmetic problem
PostPosted: Sat, 28 Jan 2012 08:38:06 UTC 
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selfluminous wrote:
prove that if 1+2^n+4^n is a prime number, then n is a power of 3

im not sure where to start with this kind of problem


Hint:

1+2^n+4^n={8^n-1\over 2^n-1

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selfluminous
 Post subject: Re: arithmetic problem
PostPosted: Sat, 28 Jan 2012 10:26:12 UTC 
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Shadow wrote:
selfluminous wrote:
prove that if 1+2^n+4^n is a prime number, then n is a power of 3

im not sure where to start with this kind of problem


Hint:

1+2^n+4^n={8^n-1\over 2^n-1

oh damn it :shock: i got to that point before, but didn't found the solution. Now that you suggest it, i will try again
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selfluminous
 Post subject: Re: arithmetic problem
PostPosted: Sat, 28 Jan 2012 16:00:53 UTC 
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selfluminous wrote:
Shadow wrote:
selfluminous wrote:
prove that if 1+2^n+4^n is a prime number, then n is a power of 3

im not sure where to start with this kind of problem


Hint:

1+2^n+4^n={8^n-1\over 2^n-1

oh damn it :shock: i got to that point before, but didn't found the solution. Now that you suggest it, i will try again

doesnt work
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outermeasure
 Post subject: Re: arithmetic problem
PostPosted: Sat, 28 Jan 2012 16:26:02 UTC 
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selfluminous wrote:
selfluminous wrote:
Shadow wrote:
selfluminous wrote:
prove that if 1+2^n+4^n is a prime number, then n is a power of 3

im not sure where to start with this kind of problem


Hint:

1+2^n+4^n={8^n-1\over 2^n-1

oh damn it :shock: i got to that point before, but didn't found the solution. Now that you suggest it, i will try again

doesnt work


What doesn't work? It works exactly as intended.

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Shadow
 Post subject: Re: arithmetic problem
PostPosted: Sat, 28 Jan 2012 18:06:15 UTC 
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selfluminous wrote:
selfluminous wrote:
Shadow wrote:
selfluminous wrote:
prove that if 1+2^n+4^n is a prime number, then n is a power of 3

im not sure where to start with this kind of problem


Hint:

1+2^n+4^n={8^n-1\over 2^n-1

oh damn it :shock: i got to that point before, but didn't found the solution. Now that you suggest it, i will try again

doesnt work


I'm pretty sure it works.

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selfluminous
 Post subject: Re: arithmetic problem
PostPosted: Sat, 28 Jan 2012 18:09:31 UTC 
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sorry i mean from 1+2^n+4^n = {8^n-1\over 2^n -1
i got 1+2^n+4^n = {7(2^{3(n-1)}+2^{3(n-2)}+...+1)\over 2^{n -1}+...+1
hmm i have tried some values of n 1+2^n+4^n seems to be divisible by 7 when n is not divisible by 3
how to connect those clues -_-
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Shadow
 Post subject: Re: arithmetic problem
PostPosted: Sat, 28 Jan 2012 18:22:19 UTC 
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selfluminous wrote:
sorry i mean from 1+2^n+4^n = {8^n-1\over 2^n -1
i got 1+2^n+4^n = {7(2^{3(n-1)}+2^{3(n-2)}+...+1)\over 2^{n -1}+...+1
hmm i have tried some values of n 1+2^n+4^n seems to be divisible by 7 when n is not divisible by 3
how to connect those clues -_-


Try treating this as a question about polynomials instead of numbers, when does (x^m-1)|(x^n-1)?

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selfluminous
 Post subject: Re: arithmetic problem
PostPosted: Sat, 28 Jan 2012 18:45:36 UTC 
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(x^m-1)|(x^n-1) when m|n
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Shadow
 Post subject: Re: arithmetic problem
PostPosted: Sun, 29 Jan 2012 18:57:10 UTC 
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selfluminous wrote:
(x^m-1)|(x^n-1) when m|n


Perfect.

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