As we all know, if $a_0+a_1x+...+a_nx^n=0$ has a solution of $p+qi$ where $\left\{ a_0,a_1,...,a_n \right\} \subset \mathbb{R}\$, then $p-qi$ is also a solution.

Similarly, if $a_0+a_1x+...+a_nx^n=0$ has a solution of $p+q\sqrt { m }$ where $\left\{ a_{ 0 },a_{ 1 },...,a_{ n } \right\} \subset \mathbb{Q}\ \quad and\quad \sqrt { m } \in \mathbb{I}\$, then $p-q\sqrt { m }$ is also a solution.

I was playing around with it until I suddenly got curious. If $a_{ 0 }+a_{ 1 }x+...+a_{ n }x^{ n }=0\quad (n\ge 4\quad and\quad n\in \mathbb{N}\ )$ has a solution of $p\sqrt { m } +qi\quad (\sqrt { m } \in \mathbb{I}\ )$, then is it safe to assume that it also has the solutions of $p\sqrt { m } -qi,\quad -p\sqrt { m } +qi,\quad -p\sqrt { m } -qi\quad$?

I tried it for $n=4$, and it worked.

Can someone confirm this for all values of n's?

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## Comments

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TopNewestYes, your observation is correct. Can you figure out how to prove it?

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Let $p + q\sqrt{m}$ be a root of a polynomial $f(x)$.

We can write $f(x)$ = $(x-(p + q\sqrt{m})$$(x-(p - q\sqrt{m}))$$a(x)$ + $r(x)$

$r(x)$'s degree would be 1,

$\implies r(x) = ax + b$

If we get $a=b=0$,then $p - q\sqrt{m}$ will also be root of $f(x)$

Put $x = p + q\sqrt{m}$ & get $a=b=0$.

(This is not a proof but a hint.)

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