
Re: How to solve diophantine equation x1^2 + x2^2 + ...+ x5^2 = x6^2
Posted:
Apr 25, 2012 3:21 PM


> > 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 > nonzero 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.

