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.
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.
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.)