Solve $ x^2+y^2=4, z^2+t^2=9, xt+yz=6 $ in integers 
find answers of this system of equations in integers$$
\left\{ 
\begin{array}{c}
x^2+y^2=4 \\ 
z^2+t^2=9 \\ 
xt+yz=6
\end{array}
\right. 
$$

things I have done: we can observe that $$(x^2+y^2)(z^2+t^2)=(xt+yz)^2+(xz-yt)^2 \rightarrow xz-yt=0$$ 
summing up this with first and second equality $$(x+z)^2+(y-t)^2=13$$
at this stage I used guessing answers. putting like $x= 0,z=3,y=2,t=0$.is there a better way to doing this without guessing and making sure that all answer found?   
 A: Perhaps you are working too hard. The solutions of the first equation are $x=\pm 2$, $y=0$, or the opposite. The solutions of $z^2+t^2=9$ are equally simple. 
Now look at the third equation.  
A: Letting $u=(x,y)$ and $v=(z,t)$ if $\theta$ is the angle between these two vectors, then:
$$\cos\theta = \frac{u\cdot v}{|u||v|} = \frac{xz+yt}{\sqrt{x^2+y^2}\sqrt{z^2+t^2}}=1$$
So the vectors point in the same direction, and the only real solutions are when $(z,t)=\frac{3}{2}(x,y)$. Then you can try the integer solutions for $x^2+y^2=4$ to find all solutions.
A: In this case I would suggest that guessing is in fact the best way, as so few guesses are needed.  (I would not suggest this if the numbers on the RHSs were much bigger.)  From $x^2+y^2=4$ we have
$$(x,y)=(2,0)\,,\ (-2,0)\,,\ (0,2)\,,\ (0,-2)\,,$$
similarly there are four possibilities for $(z,t)$, and you can quite quickly find which of the combinations satisfy the third equation:


*

*if $x=0$ then $yz=6$ so $y=\pm2$, $z=\pm3$ and $t=0$;

*if $x\ne0$ then $y=0$ and something similar happens.

