<s>[INST] 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 $\mathbb{Q}(\sqrt{5}).$
 [/INST] From your assumption 
$$(\sqrt5-a)^2 =\sqrt2b^2+2c^2+2d^2\sqrt2 +2bc\sqrt[4]{8}+2cd\sqrt[4]{32}+4bd$$ which after rearrangement is  $2c^2+4bd  +(b^2+2d^2+4cd)\sqrt2+2\sqrt2bc\sqrt[4]2$.
So multiplying by $\sqrt2$ we get  $$\sqrt2(\sqrt5-a)^2 = 2(b^2+2d^2+4cd)+(2c^2+4bd)\sqrt2 + 4bc\sqrt[4]2$$  Denoting the integers $2(b^2+2d^2+4cd),\ (2c^2+4bd)$ by $m$ and $n$ respectively bringing them to the right and squaring again we get 
$$\bigg[\sqrt2[(\sqrt5-a)^2- n]-m\bigg]^2=16b^2c^2\sqrt2 $$
Now the problem is simpler as we have got rid of all 4th roots and everything is  a square root in the above. Getting a contradiction from here should be  manageable.
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