Hi,

I am not sure if I am approaching the problem below in the right way. Is there any chance that someone can check it for me,please?

Show that the equation

$\displaystyle x^2=3(y^2 + z^2) has \ no \ zero \ integer \ solution.$

I assumed by contradiction that $\displaystyle (x,y,z)$ is an integer solution to our problem and it has the smallest value of $\displaystyle z$ of any such solution.

We can also note that $\displaystyle (x, \sqrt{3}y, \sqrt{3}z)$ is in fact a Pythagoran triple as $\displaystyle 3y^2+3z^2=x^2$.

Suppose that

$\displaystyle \sqrt{3}y=2st\\ \sqrt{3}z=s^2+t^2\\x=s^2+t^2$

for some co-prime integers $\displaystyle s \ and \ t$. Let $\displaystyle s$ be even and $\displaystyle t$ be odd, but that also imlpies that either $\displaystyle s$ or $\displaystyle t$ is divisible by$\displaystyle \sqrt{3}$. Hence contradict the fact that they are integer.

Thank you