Just Curious

As we all know, if a0+a1x+...+anxn=0a_0+a_1x+...+a_nx^n=0 has a solution of p+qip+qi where {a0,a1,...,an}R \left\{ a_0,a_1,...,a_n \right\} \subset \mathbb{R}\ , then pqip-qi is also a solution.

Similarly, if a0+a1x+...+anxn=0a_0+a_1x+...+a_nx^n=0 has a solution of p+qmp+q\sqrt { m } where {a0,a1,...,an}Q andmI \left\{ a_{ 0 },a_{ 1 },...,a_{ n } \right\} \subset \mathbb{Q}\ \quad and\quad \sqrt { m } \in \mathbb{I}\ , then pqmp-q\sqrt { m } is also a solution.

I was playing around with it until I suddenly got curious. If a0+a1x+...+anxn=0(n4andnN )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 pm+qi(mI )p\sqrt { m } +qi\quad (\sqrt { m } \in \mathbb{I}\ ), then is it safe to assume that it also has the solutions of pmqi,pm+qi,pmqip\sqrt { m } -qi,\quad -p\sqrt { m } +qi,\quad -p\sqrt { m } -qi\quad ?

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

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

Note by Nick Lee
5 years, 8 months ago

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

Calvin Lin Staff - 5 years, 7 months ago

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

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

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

    r(x)=ax+b\implies r(x) = ax + b

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

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

(This is not a proof but a hint.)

Harsh Shrivastava - 5 years, 7 months ago

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대입해서 복소수상등 무리수상등을 동시해하시면되는데 그게 힘들죠... 증명해보고잇어요!

찬홍 민 - 5 years, 6 months ago

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