using the primitive element of biquadratic fieldUsingThePrimitiveElementOfBiquadraticField2008-03-13 06:43:092008-03-15 13:35:20Applicationprimitive element of biquadratic field% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}
Let $m$ and $n$ be two distinct squarefree integers $\neq 1$.\, We want to express their square roots as polynomials of
\begin{align}
\alpha\, := \,\sqrt{m}\!+\!\sqrt{n}
\end{align}
with rational coefficients.\\
If $\alpha$ is \PMlinkname{cubed}{CubeOfANumber}, the result \PMlinkescapetext{contains} no terms with $\sqrt{mn}$:
\begin{align*}
\alpha^3 &= (\sqrt{m})^2+3(\sqrt{m})^2\sqrt{n}+3\sqrt{m}(\sqrt{n})^2+(\sqrt{n})^3\\
&= m\sqrt{m}+3m\sqrt{n}+3n\sqrt{m}+n\sqrt{n}\\
&= (m+3n)\sqrt{m}+(3m+n)\sqrt{n}
\end{align*}
Thus, if we subtract from this the product $(3m\!+\!n)\alpha$, the $\sqrt{n}$ term vanishes:
$$\alpha^3-(3m+n)\alpha = (-2m+2n)\sqrt{m}$$
Dividing this equation by $-2m\!+\!2n$ ($\neq 0$) yields
\begin{align}
\sqrt{m}\, = \,\frac{\alpha^3-(3m+n)\alpha}{2(-m+n)}.
\end{align}
Similarly, we have
\begin{align}
\sqrt{n}\, = \,\frac{\alpha^3-(m+3n)\alpha}{2(m-n)}.
\end{align}
The \PMlinkescapetext{representations} (2) and (3) may be interpreted as such polynomials as intended.
Multiplying the equations (2) and (3) we obtain a corresponding \PMlinkescapetext{representation} for the square root of $mn$ which also lies in the quartic field \,$\mathbb{Q}(\sqrt{m},\,\sqrt{n}) = \mathbb{Q}(\sqrt{m}\!+\!\sqrt{n})$:
\begin{align*}
\sqrt{mn}\, = \,\frac{\alpha^6-4(m+n)\alpha^4+(3m^2+10mn+3n^2)\alpha^2}{4(-m^2+2mn-n^2)}\\
\end{align*}
For example, in the special case \,$m := 2,\; n := 3$\, we have
$$\sqrt{2} = \frac{\alpha^3-9\alpha}{2}, \quad \sqrt{3} = -\frac{\alpha^3-11\alpha}{2}, \quad
\sqrt{6} = \frac{-\alpha^6+20\alpha^4-99\alpha^2}{4}.\\$$
\textbf{Remark.}\, The sum (1) of two square roots of positive squarefree integers is always irrational, since in the contrary case, the equation (3) would say that $\sqrt{n}$ would be rational; this has been proven impossible \PMlinkname{here}{SquareRootOf2IsIrrationalProof}.