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# using the primitive element of biquadratic field

Let $m$ and $n$ be two distinct squarefree integers $\neq 1$. We want to express their square roots as polynomials of

$\displaystyle\alpha\;:=\;\sqrt{m}\!+\!\sqrt{n}$ | (1) |

with rational coefficients.

If $\alpha$ is cubed, the result contains no terms with $\sqrt{mn}$:

$\displaystyle\alpha^{3}$ | $\displaystyle\;=\;(\sqrt{m})^{3}+3(\sqrt{m})^{2}\sqrt{n}+3\sqrt{m}(\sqrt{n})^{% 2}+(\sqrt{n})^{3}$ | ||

$\displaystyle\;=\;m\sqrt{m}+3m\sqrt{n}+3n\sqrt{m}+n\sqrt{n}$ | |||

$\displaystyle\;=\;(m+3n)\sqrt{m}+(3m+n)\sqrt{n}$ |

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

$\displaystyle\sqrt{m}\;=\;\frac{\alpha^{3}\!-\!(3m\!+\!n)\alpha}{2(-m\!+\!n)}.$ | (2) |

Similarly, we have

$\displaystyle\sqrt{n}\;=\;\frac{\alpha^{3}\!-\!(m\!+\!3n)\alpha}{2(m\!-\!n)}.$ | (3) |

The representations (2) and (3) may be interpreted as such polynomials as intended.

Multiplying the equations (2) and (3) we obtain a corresponding representation for the square root of $mn$ which also lies in the quartic field $\mathbb{Q}(\sqrt{m},\,\sqrt{n})=\mathbb{Q}(\sqrt{m}\!+\!\sqrt{n})$:

$\displaystyle\sqrt{mn}\;=\;\frac{\alpha^{6}\!-\!4(m\!+\!n)\alpha^{4}\!+\!(3m^{% 2}\!+\!10mn\!+\!3n^{2})\alpha^{2}}{4(-m^{2}\!+\!2mn\!-\!n^{2})}$ |

For example, in the special case $m:=2,\;n:=3$ we have

$\sqrt{2}\;=\;\frac{\alpha^{3}\!-\!9\alpha}{2},\quad\sqrt{3}\;=\;-\frac{\alpha^% {3}\!-\!11\alpha}{2},\quad\sqrt{6}\;=\;\frac{-\alpha^{6}\!+\!20\alpha^{4}\!-\!% 99\alpha^{2}}{4}.\\$ |

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 here.

## Mathematics Subject Classification

11R16*no label found*

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