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outermeasure
 Post subject: no primes
PostPosted: Wed, 18 Jan 2012 15:28:48 UTC 
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A easy one:

(1) Prove that there are no primes in the sequence 10001,100010001,1000100010001, ...
(2) Prove that, for every natural number n>1, there are only finitely many primes in the sequence 1+n^4,1+n^4+n^8,1+n^4+n^8+n^{12},\dots
(3) What is the minimum number of terms from the sequence 1+n^4,1+n^4+n^8,1+n^4+n^8+n^{12},\dots you need to test to guarantee the sequence contains no primes?

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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: no primes
PostPosted: Thu, 19 Jan 2012 10:04:39 UTC 
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There don't seem to be many takers, I hope you don't mind outermeasure, but for those mulling it over,

Hint:

Spoiler:
This really amounts to some long division, and counting becomes easy once you figure out why there are only finitely many (at least if you do it like I did).

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outermeasure
 Post subject: Re: no primes
PostPosted: Thu, 19 Jan 2012 11:06:40 UTC 
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Shadow wrote:
There don't seem to be many takers, I hope you don't mind outermeasure, but for those mulling it over,

Hint:

Spoiler:
This really amounts to some long division, and counting becomes easy once you figure out why there are only finitely many (at least if you do it like I did).


Indeed that's my approach too.

Spoiler:
It is secretly an algebra question rather than a number theory question.

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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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alstat
 Post subject: Re: no primes
PostPosted: Thu, 31 May 2012 20:48:42 UTC 
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I was working on another similar problem: trying to prove 10^k + 1 is never prime for k > 2, when I remembered this problem that I didn't have time to look at a few months ago.

To answer this, consider S_k = 1 + n^4 + n^8 + ... + n^{4(k-1)} where n > 1, k > 1.

Note that S_{km} = S_k(1 + n^{4k} + n^{8k} + ... + n^{4k(m-1)}), so S_k is composite whenever k is.

Also S_k = \frac{n^{4k} - 1}{n^4 - 1} as a geometric series with r = n^4.

So S_k = \frac{(n^k - 1)(n^k + 1)(n^{2k} + 1)}{(n - 1)(n + 1)(n^2 + 1)}.

If k is odd, this equals
(n^{k-1} + n^{k-2} + ... + 1)(n^{k-1} - n^{k-2} + ... + 1)(n^{2(k-1)} - n^{2(k-2)} + ... + 1)
and so S_k is composite when k is odd (in particular, when k is an odd prime).

So the S_k contains either one or zero primes, depending on whether S_2 = 1 + n^4 is prime.

For n = 10, S_2 = 10001 = (73)(137) is composite, so the sequence 10001, 100010001, 1000100010001, ... contains no primes.
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