Solution of a quadratic diophantine equation I try to solve the Diophantine quadratic equation:
$$X^2+Y^2+Z^2=3W^2.$$ Obviously, there is a non-trivial solution: $(1,1,1,1)$. So I tried to apply Jagy's method: Solutions to $ax^2 + by^2 = cz^2$ .
I consider integers $t,p,q,r$ and the point $P=(1,1,1)$ of the sphere $X^2+Y^2+Z^2=3$. I look for a second point on the sphere in the form $(1+pt,1+qt,1+rt)$. That gives $p^2t+q^2t+r^2t+2p+2q+2r=0$. This implies that
$$\left(1-\frac{2p(p+q+r)}{p^2+q^2+r^2},1-\frac{2q(p+q+r)}{p^2+q^2+r^2},1-\frac{2r(p+q+r)}{p^2+q^2+r^2}\right)$$
is a rational point of the sphere $X^2+Y^2+Z^2=3$.    But what to do with that? Thanks in advance. 

From above, the below mentioned equation has parametric solution,
$x^2+y^2+z^2=3w^2$
$x=k^2-6k+1$
$y=k^2-2k+5$
$z=5k^2-2k+1$
$w=3k^2-2k+3$

