Diophantine equation of second degree $x^2+y^2+z^2=2t^2$ How to solve this Diophantine equation of second degree? 
Solution, references, anything. I will be very grateful. 
$$x^2+y^2+z^2=2t^2$$
Thank you.
 A: You can start with $x,y,z$ any Pythagorean triple, then take $t=z$.  There are more.
A: You can fix $z$ at $t$, and vary $x$ and $y$,
After fixing $z$ at $t$, the equation becomes,
$$x^2+y^2=t^2\dots(1)$$
One solution of this equation is $(0,t)$
$(1)$ is the equation of a circle in $x-y$ plane with radius $t$.
Inorder to find all the solution of $(1)$ we follow the following method,
Let $(x_0,y_0)$ be some other solution,
We will draw the line connecting $(x_0,y_0)$ and $(0,t)$,
Equation of this line would be $\frac{y_0-t}{x_0}=m,m\in R$
$\Rightarrow y_0=mx_0+t\dots (2)$
Putting this in $(1)$ we have,
$x_0^2(1+m^2)+2mx_0t+t^2=t^2$
$\Rightarrow x_0^2(1+m^2)+2mx_0t=0$
$x_0(1+m^2)+2mt=0$ Considering $x_0\ne 0$ as then we get the same point$(0,t)$
$\Rightarrow x_0=-\frac{2mt}{1+m^2}$
Putting this into (2) we get,
$y_0=\frac{t(1-m^2)}{1+m^2}$
Now $(x_0,y_0)$ was any arbitrary solution of this equation.For each such solution we will get a line . And for each such line we will get another solution(quite eacy to see), which implies if we consider all $m\in R$ we will get all the solutions of (1).
Thus one set of solution of the original equation is, $(-\frac{2mt}{1+m^2},\frac{t(1-m^2)}{1+m^2},t)$ $m\in R$.  
A: Expanding on Ross Millikan's suggestion, let $m$ and $n$ be any two integers, and then let
$$x = m^2 - n^2$$
$$y = 2mn$$
$$z = m^2 + n^2$$
$$t = z$$
Then $x,y,z$ form a Pythagorean triple, which means that $x^2 + y^2 = z^2$. This gives 
$x^2 + y^2 + z^2 = 2z^2 = 2t^2$. There are other ways to generate Pythagorean triples, of course; this is just the one that came to mind.
A: It is strange that so many and so this kind of equation vozyatsya.Takogo vsegda.To elementary equations are solved there can always say under what factors can Met solutions in integers and when they are not.
Okay. equation:
$X^2+Y^2+Z^2=2T^2$
Solutions can be written:
$X=(p-q)^2+q^2-2sp$
$Y=2(q+s)(q+s-p)$
$Z=p^2+2s^2-2p(q+s)$
$T=(p-q)^2+(q+s)^2+s(s-2p)$
$p,s,q$ - any integers any sign.
A: For every $t$, $x^2+y^2+z^2=2t^2$, defines a 3-Dimensional sphere with radius $\sqrt{2}t$. Thus, every point on the surface of the sphere is a solution for the diophantine equation. If you are interested in characterizing all the solutions, this should help
\begin{align}
r&=\sqrt{2}t \\
x&=r\sin{(\theta)}\cos(\phi) \\
y&=r\sin{(\theta)}\sin(\phi) \\
z&=r\cos{(\theta)} \\
\end{align} 
where $0\leq \theta \leq \pi$ and $0 \leq \phi \leq 2\pi$. Thus, substituting any value for $t$ , $\theta$ and $\phi$ in the above mentioned equations will give you a solution. 
