
Re: How to solve diophantine equation x1^2 + x2^2 + ...+ x5^2 = x6^2
Posted:
Apr 25, 2012 1:57 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 ?
If you allow x5=0 there is x1=1 x2=1 x3=1 x4=1 x5=0 x6=2. This is primitive.
Even if you decide not to allow zero values, this example shows there can be no simple "congruence argument" for no primitive solutions of the type you mention.

