Dependence on parameters of solvability of a non-linear Diophantine system

Question

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.

Answer 1

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 $(v_1,v_2,s) = (5,-12,13)$.

Answer 2

For the system of Diophantine equations.

$$\left\{\begin{aligned}&a^2+b^2=c^2\\&ax+by=cz\end{aligned}\right.$$

You can write the parameterization of the solutions.

$$a=k^2-p^2$$

$$b=2kp$$

$$c=k^2+p^2$$

$$x=tk^2+2(s-t)kp+3tp^2$$

$$y=tk^2-2tkp+(2s-t)p^2$$

$$z=tk^2+2skp-3tp^2$$

This means that for every Pythagorean triple has infinitely many solutions.