Deriving a contradiction How can I derive a contradiction from the following nasty statement:
Assume $\sqrt{5} = a + b\sqrt[4]{2} + c\sqrt[4]{4} + d\sqrt[4]{8},$ with $a,b,c,d \in \mathbb{Q}$?
This is the last piece of an effort to prove that the polynomial $x^4-2$ is irreducible over $\mathbb{Q}(\sqrt{5}).$

Starting from 
$$\tag1m\sqrt 5=a+b\alpha+c\alpha^2+d\alpha^3 $$
(where $\alpha^4=22$) with $a,b,c,d\in\mathbb Z$  you find
$$\begin{align}5m^2&=(a+b\alpha+c\alpha^2+d\alpha^3)^2\\&=(a^2+2c^2+4bd)+(2ab+4cd)\alpha+(b^2+2d^2+2ac)\alpha^2+(2ad+2bc)\alpha^3\end{align}$$ 
If you alredy know that $1,\alpha,\alpha^2,\alpha^3$ are linearly independent over $\mathbb Q$, this implies
$$ \begin{align}a^2+2c^2+4bd-5m^2&=0\\
ab+2cd&=0\\
b^2+2d^2+2ac&=0\\
ad+bc&=0\end{align}$$
From the thirs we have immediatelythat $b$ must be even.
Assume $m$ is odd.
Then from the first, $a$ must be odd. Then the first reads $1+2c^2+0-5\equiv 0\pmod 8$, i.e., $c^2\equiv 2\pmod 4$, which is absurd.
Therefore $m$ is even. Then from the first $a$ must be even. Also from the first, $c$ must be even. And from the thord $d$ must be even.
Thus: For any solution $(m,a,b,c,d)\in\mathbb Z^5$ of $(1)$, all variables are even. But if there exists a nontrvial solution at all, then certainly there exists one with at least one variable odd, obtained by dividing out the larges common power of $2$. We conclude that $m=a=b=c=d=0$ is the only integer solution to $(1)$. Therefore, there is no rational solution for 
$$\sqrt 5=a+b\alpha+c\alpha^2+d\alpha^3 $$
