First, the problem statement : "Consider the equation $x^{2}+y^{2}-3z^{2}-3t^{2}=0$. The total number of integral solutions of this equation in the range of the first 10000 numbers, i.e., $1\leq x,y,z,t\leq10000$ is: (a.) 200 (b.) 55 (c.) 100 (d.) 1 (e.) None of the above"
I know how to find the number of integral solutions for first degree equations,i.e. form $x_{1}+x_{2}+x_{3}+...+x_{n}=0$ but I am unable to start with this equation. One observation that I made ( a very obvious one ) is that $x^{2}+y^{2}=3\left( z^{3}+t^{2}\right)$ and thus we can further have $\left(x+y\right)^{2}-3\left(z+t\right)^{2}+6zt-2xy=0$ but this doesn't seem to take me anywhere towards the number of integral solutions.
Any insights or reference that I should refer to solve the above problem?
The solution given in key is option (e.)
There are no solutions.
Assume the contrary, then by the well-ordering principle there is a solution $(x, y, z, t) $ such that $ x^2 + y^2 $ is minimized. If $ x^2 + y^2 = 3z^2 + 3t^2 $, reducing both sides modulo 3 and noting that the quadratic residues mod 3 are $ 0, 1 $ we have that both $ x $ and $ y $ are divisible by $ 3 $, so we can let $ x = 3x' $ and $ y = 3y' $ to find $ 9x'^2 + 9y'^2 = 3z^2 + 3t^2 $ or $ z^2 + t^2 = 3x'^2 + 3y'^2 $. However, then $ (z, t, x', y') $ is a solution such that $ z^2 + t^2 < x^2 + y^2 $ (assuming that $ x^2 + y^2 > 0 $), contradiction.
The only solution occurs when $ x^2 + y^2 = 0 $ or $ x=y=z=t=0 $, which does not lie in the desired interval.