Show that all solutions to x^2+y^2+z^2=w^2 with the restrictions

w=x+y,

x>0, y>0, z>0, w>0,

(x,y,z,w)=1 and

2|x (i.e. x is even)

are given by

x=2a^2,

y=b^2,

z=2ab,

w=2a^2+b^2

where a>0, b>0, (a,b)=1 and b is odd.

Printable View

- July 9th 2009, 07:34 AMdiddledabbleDiophantine Proof Help using Pythagorean Triples
Show that all solutions to x^2+y^2+z^2=w^2 with the restrictions

w=x+y,

x>0, y>0, z>0, w>0,

(x,y,z,w)=1 and

2|x (i.e. x is even)

are given by

x=2a^2,

y=b^2,

z=2ab,

w=2a^2+b^2

where a>0, b>0, (a,b)=1 and b is odd. - July 9th 2009, 07:09 PMhalbard
Firstly, , so .

Since is even let . Then , so must be even, say .

Now any common factor of and must be a factor of and of , since . But , so must hold.

Therefore also. But , so . Therefore since and we have and with .

Therefore and .

Finally , so as take with and (letting and gives the same solutions.)