I am working through some examples from Silverman's The Arithmetic of Elliptic Curves and I've run into the following system of equations, $$\left\{x^2+y^2=2,x^2+z^2=4\right\}$$ I am quite confident that it does not have rational solutions but I'm struggling to find a way to show it. Any hint, please?
Edit: This comes from trying to use the complete 2-descent procedure in AEC to the curve $y^2=x^3-4x$. The method yields this system of equations, which should not have solutions since the rank of the curve is $0$. But I don't manage to show it.
Assume the system of equations has a rational solution. Thus, $x=\frac{a}{b}$ with $\gcd(a,b)=1$, $y=\frac{c}{d}$ with $\gcd(c,d)=1$ and $z=\frac{e}{f}$ with $\gcd(e,f)=1$. We then get from $x^2+y^2=2$ that
$$\left(\frac{a}{b}\right)^2+\left(\frac{c}{d}\right)^2=2\tag{1}\label{eq1A}$$
Multiplying both sides by $b^{2}d^{2}$ gives
$$a^{2}d^{2} + c^{2}b^{2} = 2b^{2}d^{2} \;\;\to\;\; a^{2}d^{2} = b^{2}(2d^{2}-c^{2}) \tag{2}\label{eq2A}$$
Thus, $b^{2} \mid a^{2}d^{2}$. Since $\gcd(a,b)=1$, then $b^{2}\mid d^{2}$. Similarly, from \eqref{eq2A} we have $c^{2}b^{2}=d^2(2b^2-a^2)$, which leads to $d^2\mid b^2$. This means $b^2 = d^2$. Dividing both sides of \eqref{eq2A} by $d^2$, we get
$$a^2 + c^2 = 2b^2 \tag{3}\label{eq3A}$$
Using a similar procedure with the second equation of $x^2 + z^2 = 4$ gives that $b^2 = f^2$ and
$$a^2 + e^2 = 4b^2 \tag{4}\label{eq4A}$$
From \eqref{eq3A}, note $a$ and $c$ must have the same parity. If they are both even, then $b$ must also be even, contradicting that $\gcd(a,b)=1$. Thus, they must both be odd. In \eqref{eq4A}, this means that $e$ must be odd as well. However, then $a^2 \equiv e^2 \equiv 1 \pmod{4}$, so the LHS of \eqref{eq4A} is congruent to $2$ modulo $4$, while the RHS is congruent to $0$ modulo $4$ instead!
This contradiction proves there are no rational solutions to the $2$ sets of equations.