<s>[INST] Show that $u-v=\sqrt5$ given that $u-v>0$ given that $u=b+b^4$ and $v=b^2+b^3$ We're also given that $b$ is a root of $z^5-1=0$ $b^4+b^3+b^2+b+1=0$
If $u=b+b^4$ and $v=b^2+b^3$, show that
i) $u+v=uv=-1$
ii) $u-v=\sqrt5$ given that $u-v>0$
I managed to do part i):
Plugging in $u$ and $v$: $(b+b^4)+(b^2+b^3)=-1$ (using $b^4+b^3+b^2+b+1=0$)
$uv=(b+b^4)(b^2+b^3)$
$=(b^3+b^4+b^6+b^7)$
$=b^3(1+b+b3+b4)$
$=b^3(-b^2)$
$=-b^5$
$=-1^5$
$=-1$ (using $z^5=1$)
For ii)
$u-v=(b+b^4)-(b^2+b3)>0$
$(b+b^4)-(-b-b^4-1)>0$
$2b+2b^4+1>0$
$b+b^4>\frac{-1}{2}$
Although it's a dead-end after that.
What's the general idea for going about solving part ii)?
 [/INST] Just compute
$$
(u-v)^2=(b+b^4-b^2-b^3)^2=-(b^4+b^3+b^2+b+1)+5=5.
$$
We have used $b^5=1$ in the second step.
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