(a) Using modular arithmetic Show that the equation $$x^2+y^2+z^2=2xyz$$ has no integer solutions except $x=y=z=0$
(b) Using the results in part (a), show that the equation $$x^3+2y^3+4z^3=0$$ has no integer solutions except $x=y=z=0$
What i tried
For part (a),
At least one of x,y,z is odd. Can it be even? Reduce mod 4
$(even)^2=0 (mod 4)$ $(odd)^2=1 (mod 4)$
implies that LHS = 2 (mod 4)
all 3 are even ie, $x=2x_{1}$, $y=2y_{1}$, $z=2z_{1}$ implies that $$4x_{1}^2+4y_{1}^2+4z_{1}^2$$
(b) I have to somehow reduce the equation of part(b) to something similar to that of part(a).
Equation of part (b) $$x^3+2y^3+4z^3=0$$ removing the $x,y,z$ term, we have $$x^2+2y^2+4z^2=0$$ and since $x^2$ is even, $x$ is even we have $x=2a$, doing the same for $y$, $y=2b$ and $z$, $z=2c$ we have,the equation
$$(2a)^2+2*(2b)^2+4*(2c)^2=0$$, However im unsure of how to continue from here, could anyone help me with this question, especially part (b). Thanks
