MHB How Can I Derive a Contradiction from This Mathematical Statement?

[kalish1]

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 $\mathbb{Q}(\sqrt{5}).$*

 

[Kiwi1]

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 $\mathbb{Q}(\sqrt{5}).$*

Since no one has replied I'll give it a crack (at the risk of leading you astray).

Let's multiply your equation through by the least common multiple of the denominators of a,b,c and d. I'll call that number e. Then we have:

$e \sqrt{5} = a' + b'\sqrt[4]{2} + c'\sqrt[4]{4} + d'\sqrt[4]{8},$ with $a',b',c',d',e \in \mathbb{Z}$

Let's define $x=\sqrt[4]{2}$

so:
$e \sqrt{5} = a' + b'x + c'x^2 + d'x^3,$ with $a',b',c',d',e \in \mathbb{Z}$

squaring both sides and defining a",b",c",d",e" appropriately:
$5e''= a''+ b''x + c''x^2 + d''x^3,$ with $a'',b'',c'',d'',e'' \in \mathbb{Z}$

or
$a'''+ b''x + c''x^2 + d''x^3=0$, with $a''',b'',c'',d'' \in \mathbb{Z}$ and $x \not\in \mathbb{Q}$

It seems to me it must be possible to disprove this last statement. But I can't see how.