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 Post subject: Divisibility Question
PostPosted: Sat, 12 May 2012 23:08:04 UTC 
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Find all positive integers n \ge 3, such that:

n \mid 2^n-7

My Work:

I tried a lot of numbers.~

Clearly, n is odd and not divisible by 7. Based on the numbers that I have tried I would like to conjecture that the only number is n=5. Also, by Euler's Theorem, we note that the only prime that trivially satisfies this is n=5. But, I don't see any way to show this for composite numbers.

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PostPosted: Sat, 12 May 2012 23:42:15 UTC 
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rdj5933mile5math64 wrote:
Find all positive integers n \ge 3, such that:

n \mid 2^n-7

My Work:

I tried a lot of numbers.~

Clearly, n is odd and not divisible by 7. Based on the numbers that I have tried I would like to conjecture that the only number is n=5. Also, by Euler's Theorem, we note that the only prime that trivially satisfies this is n=5. But, I don't see any way to show this for composite numbers.


By the CRT you only need to check on prime powers, so you want to see for what prime powers, 2^{p^k}\equiv 7\mod p^k, but you know that \varphi(p^k)=p^k-p^{k-1}, so you're saying that 7\equiv 2^{p^k}\cdot 2^{-p^{k-1}}\cdot 2^{p^{k-1}}\mod p^k

In other words,

7^{p}\equiv 7\mod p^k, i.e. 7^{p-1}\equiv 1\mod p^k.

Try thinking about this.

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PostPosted: Sun, 13 May 2012 10:56:44 UTC 
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Shadow wrote:
rdj5933mile5math64 wrote:
Find all positive integers n \ge 3, such that:

n \mid 2^n-7

My Work:

I tried a lot of numbers.~

Clearly, n is odd and not divisible by 7. Based on the numbers that I have tried I would like to conjecture that the only number is n=5. Also, by Euler's Theorem, we note that the only prime that trivially satisfies this is n=5. But, I don't see any way to show this for composite numbers.


By the CRT you only need to check on prime powers, so you want to see for what prime powers, 2^{p^k}\equiv 7\mod p^k, but you know that \varphi(p^k)=p^k-p^{k-1}, so you're saying that 7\equiv 2^{p^k}\cdot 2^{-p^{k-1}}\cdot 2^{p^{k-1}}\mod p^k

In other words,

7^{p}\equiv 7\mod p^k, i.e. 7^{p-1}\equiv 1\mod p^k.

Try thinking about this.


Hmm... p^kN\mid 2^{p^kp'}-7 does not give 2^{p^k}\equiv 7\pmod{p^k}, but (2^N)^{p^k}\equiv 7\pmod{p^k}.

Anyway, the conjecture is obviously false with n=25.

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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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PostPosted: Mon, 14 May 2012 15:08:31 UTC 
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outermeasure wrote:
Shadow wrote:
rdj5933mile5math64 wrote:
Find all positive integers n \ge 3, such that:

n \mid 2^n-7

My Work:

I tried a lot of numbers.~

Clearly, n is odd and not divisible by 7. Based on the numbers that I have tried I would like to conjecture that the only number is n=5. Also, by Euler's Theorem, we note that the only prime that trivially satisfies this is n=5. But, I don't see any way to show this for composite numbers.


By the CRT you only need to check on prime powers, so you want to see for what prime powers, 2^{p^k}\equiv 7\mod p^k, but you know that \varphi(p^k)=p^k-p^{k-1}, so you're saying that 7\equiv 2^{p^k}\cdot 2^{-p^{k-1}}\cdot 2^{p^{k-1}}\mod p^k

In other words,

7^{p}\equiv 7\mod p^k, i.e. 7^{p-1}\equiv 1\mod p^k.

Try thinking about this.


Hmm... p^kN\mid 2^{p^kp'}-7 does not give 2^{p^k}\equiv 7\pmod{p^k}, but (2^{p'})^{p^k}\equiv 7\pmod{p^k}.

Anyway, the conjecture is obviously false with n=25.


Whoops \varphi(25) \ne 8. (I did \varphi(p^k)=k(p-1) but just for n=25... ) But n=125 doesn't work.

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