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PostPosted: Tue, 11 May 2010 23:15:49 UTC 
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The expression n^2-n+nk^2 is perfect square for every value of the positive integer number n. Find k.


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PostPosted: Wed, 12 May 2010 02:42:29 UTC 
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1.

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I'm just an imagination of your figment...


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PostPosted: Wed, 12 May 2010 15:26:40 UTC 
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mathemagics wrote:
The expression n^2-n+nk^2 is perfect square for every value of the positive integer number n. Find k.


Denis wrote:
1.

... or -1.

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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: Wed, 12 May 2010 21:11:28 UTC 
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It's obvious that the solutions are the numbers \pm 1. Could you please give a full proof?


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PostPosted: Thu, 13 May 2010 18:06:36 UTC 
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mathemagics wrote:
It's obvious that the solutions are the numbers \pm 1. Could you please give a full proof?


Spoiler:
n^2-n+nk^2=n(n+k^2-1). Make n>\lvert k^2-1\rvert a prime and you get ....

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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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