I was asked to show that the following system has no rational solutions:
$y^2 = 17 + 2x^2 \\ y^2 = 34 +z^2 $
That is to say, that there are no $x,y,z \in \mathbb{Q}$ such that the above relations hold.
My attempt:
It was fairly easy to show that there are no solutions over $\mathbb{Z}$ and this is where I pretty much got stuck.
ANY help would be appreciated
$y^2 = 17 + 2x^2 \\ y^2 = 34 +z^2 $
Let $v$ be a common denominator, so that $vy, vx, vz$ are integers. Name $u = vy, $ then $w_1 = vx$ and $w_2 = vz.$ So far, I have $u^2 - 17 v^2 = 2 w_1^2$ while $u^2 - 34 v^2 = w_2^2$
Here we are assuming that $u,v$ are not both zero! If there is a common divisor of $u,v$ then it also divides $w_1, w_2$ and we may divide through by that, achieving coprime $u,v$
Now, have $\gcd(u,v) = 1$ and $$u^2 - 17 v^2 = 2 w_1^2 \; , \; \; \; u^2 - 34 v^2 = w_2^2$$
We know at least one of $u,v$ is odd. Since $u^2 - 17 v^2$ must be even, we see that $u,v$ are both odd.
However, $$ u^2 - 34 v^2 \equiv 1 - 2 \equiv 3 \pmod 4 $$ and cannot be a square.