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PostPosted: Wed, 22 Feb 2012 03:30:30 UTC 
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For what integers x,y is the equation y^2+4=x^3 satisfied?

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PostPosted: Wed, 22 Feb 2012 03:55:29 UTC 
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Shadow wrote:
For what integers x,y is the equation y^2+4=x^3 satisfied?


I know this is not what you are looking for, but y=11 and x=5 work.
As far as a general solution, I have no idea. :confused:

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PostPosted: Wed, 22 Feb 2012 03:56:50 UTC 
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Justin wrote:
Shadow wrote:
For what integers x,y is the equation y^2+4=x^3 satisfied?


I know this is not what you are looking for, but y=11 and x=5 work.
As far as a general solution, I have no idea. :confused:


I'll give you a hint: there are only finitely many solutions and more than you have already posted. :)

Also, note you have a square for y, so you may as well add 5 and -11 to up your nonexistent "score" ;)

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PostPosted: Wed, 22 Feb 2012 04:25:27 UTC 
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Another hint:
Spoiler:
\mathbb{Z}[\sqrt{-1}]=\mathcal{O}_{\mathbb{Q}(\sqrt{-1})} is Euclidean, and all units are cubes.

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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, 22 Feb 2012 04:55:44 UTC 
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Spoiler:
Hmmm...I didn't think about using Gaussian integers.
I don't get the hint about "units being cubes" though.
I will have to ponder this some more...obviously, you are prompting me
to use some clever factorization trick here(?)

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PostPosted: Wed, 22 Feb 2012 05:22:51 UTC 
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That is to absorb the unit(s) that arise when you invoke unique factorisation.

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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, 22 Feb 2012 15:20:30 UTC 
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Justin: You'll find that most diophantine equations require at least some machinery from something a bit more potent than elementary methods to solve as completely as necessary. This one is about as simple as it gets without being essentially trivial, which is why I chose it, since it's still a bit involved. :)

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