# Diophantine Proof Help using Pythagorean Triples

• Jul 9th 2009, 07:34 AM
diddledabble
Diophantine 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.
• Jul 9th 2009, 07:09 PM
halbard
Firstly, \$\displaystyle w^2=x^2+y^2+2xy\$, so \$\displaystyle z^2=2xy\$.

Since \$\displaystyle x\$ is even let \$\displaystyle x=2k\$. Then \$\displaystyle z^2=4ky\$, so \$\displaystyle z\$ must be even, say \$\displaystyle z=2m\$.

Now any common factor of \$\displaystyle x\$ and \$\displaystyle y\$ must be a factor of \$\displaystyle w=x+y\$ and of \$\displaystyle z\$, since \$\displaystyle z^2=2xy\$. But \$\displaystyle (x,y,w,z)=1\$, so \$\displaystyle (x,y)=1\$ must hold.

Therefore \$\displaystyle (k,y)=1\$ also. But \$\displaystyle 4ky=z^2=4m^2\$, so \$\displaystyle ky=m^2\$. Therefore since \$\displaystyle y>0\$ and \$\displaystyle (k,y)=1\$ we have \$\displaystyle k=a^2\$ and \$\displaystyle y=b^2\$ with \$\displaystyle (a,b)=1\$.

Therefore \$\displaystyle x=2k=2a^2\$ and \$\displaystyle w=x+y=2a^2+b^2\$.

Finally \$\displaystyle z^2=4ky=4a^2b^2\$, so as \$\displaystyle z>0\$ take \$\displaystyle z=2ab\$ with \$\displaystyle a>0\$ and \$\displaystyle b>0\$ (letting \$\displaystyle a<0\$ and \$\displaystyle b<0\$ gives the same solutions.)