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   Author  Topic: quadruple  (Read 2291 times)
Christine
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quadruple  
« on: Sep 15th, 2013, 4:56pm »
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Diophantus quadruple (1/16, 33/16, 17/4, 105/16)
 
the product of any two distinct terms + 1 is a square
 
How can you find other sets of rational quadruple?
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SEMfuze
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Re: quadruple  
« Reply #1 on: Sep 27th, 2013, 7:20am »
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y
2 = (a1x + n)(a2x + n)(a3x + n)(a4x + n)(a5x + n)
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Christine
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Re: quadruple  
« Reply #2 on: Sep 27th, 2013, 9:40am »
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on Sep 27th, 2013, 7:20am, SEMfuze wrote:
y
2 = (a1x + n)(a2x + n)(a3x + n)(a4x + n)(a5x + n)

 
It doesn't look right. Maybe I miss something here.
Please clarify.
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towr
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Re: quadruple  
« Reply #3 on: Sep 27th, 2013, 2:35pm »
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I think he meant to copy/paste the formula for a hyperelliptic curve y2 = (a1x + 1)(a2x + 1)(a3x + n)(a4x + 1)(a5x + 1)  (n=1 in this case). Equation 2 on page 2 of http://arxiv.org/pdf/math/0002088.pdf, for example (I haven't read it, and not sure how much I'd understand of it if I would).
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Christine
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Re: quadruple  
« Reply #4 on: Sep 27th, 2013, 5:15pm »
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on Sep 27th, 2013, 2:35pm, towr wrote:
I think he meant to copy/paste the formula for a hyperelliptic curve y2 = (a1x + 1)(a2x + 1)(a3x + n)(a4x + 1)(a5x + 1)  (n=1 in this case). Equation 2 on page 2 of http://arxiv.org/pdf/math/0002088.pdf, for example (I haven't read it, and not sure how much I'd understand of it if I would).

 
Ouch! Thanks. The stuff on arxiv  is tough to understand. The question I posted is more difficult than I anticipated.
 
 
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