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Topic: How to solve diophantine equation x1^2 + x2^2 + ...+ x5^2 = x6^2
Replies: 9   Last Post: Apr 25, 2012 8:21 PM

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Dan Cass

Posts: 442
Registered: 12/6/04
Re: How to solve diophantine equation x1^2 + x2^2 + ...+ x5^2 = x6^2
Posted: Apr 25, 2012 3:21 PM
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> > There is of course a primitive solution
> > (in my ordering of notation)
> > (1,1,1,1,0,2)
> > representing that
> > 1^2 + 1^2 + 1^2 + 1^2 + 0^2 = 2^2.
> >
> > This tuple is in my notation
> > (1,1,1,1,0,2)
> > and is also primitive.
> > It would be obtained using the matrix as described

> in
> > the paper, beginning with (1,0,0,0,0,1).
> >
> > It seems your use of four 1's would need a 2^2
> > somewhere.

>
> Is there any primitive solution in which all are
> non-zero and x1,x2,x3,x4 odd and x5,x6 even ?


Here's one:
x1=1 x2=3 x3=5 x4=7 x5=20 x6=22.

These seem easy to find by choosing the first four
virtually at random, and then assuming the final two
x5 and x6 to be n and n+2, so that x6^2 - x5^2 = 4n+4.



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