Diophantine equation of second degree $x^2+y^2+z^2=2t^2$

Question

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.

Answer 1

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.

Answer 2

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, $(-\frac{2mt}{1+m^2},\frac{t(1-m^2)}{1+m^2},t)$ $m\in R$.

Answer 3

You can start with $x,y,z$ any Pythagorean triple, then take $t=z$. There are more.

Answer 4

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.