For which $(k,t)\in\mathbb Z^2\times\mathbb Z$ does there exist $(v,s)\in\mathbb Z^2\times\mathbb Z$ so that $|v|^2=s^2\neq0$ and $v\cdot k+st=0$?
I do not care what the solutions $(v,s)$ are but only whether they exist or not.


This is what I have found out about existence of solutions $(v,s)\in\mathbb Z^2\times\mathbb Z$ to the system
$$
\begin{cases}
v\_1^2+v\_2^2=s^2\neq0\\
v\_1k\_1+v\_2k\_2+st=0
\end{cases}
$$
in terms of $(k\_1,k\_2,t)$:


* If $t=0$ and $k=0$, there is a solution.
* If $|t|=|k\_1|$ (or $|t|=|k\_2|$), there is a solution: Take $s=-t$ and $v=(k\_1,0)$ (or $(0,k\_2)$).
* If $t=0$ and $|k|$ is integer, there is a solution: Take $s=|k|$ and $v=(k\_2,-k\_1)$.
* If $t^2=k\_1^2+k\_2^2$, there is a solution: Take $s=-t$ and $v=k$.
* If $t^2>k\_1^2+k\_2^2$, there is no solution: Suppose there is a solution $(v,s)$. Then $s=at$ for some $a\in\mathbb Q\setminus\{0\}$. Then
$$
|t|^2=|a^{-1}st|=|a|^{-1}|v\cdot k|\leq|a|^{-1}|v||k|=|a|^{-1}|s||k|=|t||k|
$$
and so $|t|\leq|k|$.
* I have looked at some examples not covered by the previous ones using Mathematica.
These values of of $(k,t)$ give solutions: $(0,5;3)$, $(0,13;5)$, $(0,13;12)$.
These do not: $(2,4;1)$, $(3,4;2)$, $(3,5;2)$, $(4,5;3)$, $(2,3;1)$, $(0,4;2)$, $(0,5;2)$.


These observations lead me to a conjecture, but I am far from convinced about it or able to prove it:



> 
> There is a solution for $(k,t)\in\mathbb Z^2\times\mathbb Z$ if and only if at least one of the following holds:
> 
> 
> * $t=0$ and $|k|$ is integer.
> * $t^2=k\_1^2+k\_2^2$.
> * $k\_1=0$ and $\sqrt{k\_2^2-t^2}$ is integer.
> * $k\_2=0$ and $\sqrt{k\_1^2-t^2}$ is integer.
> 
> 
> 


The reason I want to know this is that a function on the torus $\mathbb T^3=\mathbb R^3/\mathbb Z^3$ turns out to satisfy a property I am interested in if and only if the Fourier transform vanishes for all $(k,t)\in\mathbb Z^2\times\mathbb Z=\mathbb Z^3$ that satisfy the above property.
Unfortunately I have no experience with non-linear Diophantine systems and my searches have not lead to much insight.




Dependence on parameters of solvability of a non-linear Diophantine system
Of course if $(v\_1, v\_2, s)$ is a solution, so is any integer multiple of it, so we may assume wlog a primitive solution: $\gcd(v\_1,v\_2,s)= 1$. In a primitive solution of $v\_1^2 + v\_2^2 = s^2$, $s$ is odd and one of $v\_1$ and $v\_2$ is odd: WLOG $v\_1$ is odd. Then for some odd $a,b$, 
$$ v\_1 = a b, \; v\_2 = \dfrac{a^2 - b^2}{2},\; s = \dfrac{a^2+b^2}{2}$$  

Now your second equation becomes a quadratic form:
$$ a b k\_1 + a^2 (t + k\_2)/2 + b^2 (t - k\_2)/2 = 0 $$
If $t = k\_2$, this has solutions with $$\dfrac{b}{a} = - \dfrac{k\_2}{k\_1}$$
Otherwise the solutions are 
$$ \dfrac{b}{a} = \dfrac{-k\_1 \pm \sqrt{k\_1^2 + k\_2^2 - t^2}}{k\_2 - t} $$
where $k\_1^2 + k\_2^2 - t^2$ must be a square in order for these to be rational.


For example, a counterexample to your conjecture is $k\_1 = 7,\; k\_2 = 4,\; t = 1$ where $(a,b) = (-3,1)$ produces the solution $(v\_1,v\_2,s) = (-3,4,5)$ and $(a,b) = (1,5)$ produces the solution $(v\_1,v\_2,s) = (5,-12,13)$.


