TITLE: If $a+\sqrt{b}$ is a root of a polynomial equation with integer coefficients, so is $a-\sqrt{b}$
QUESTION [8 upvotes]: I tried to use the Briot-Ruffini method but it didn't work.
The question I need help is:
"Prove that, if a polynomial equation with integer coefficients has the irrational number $a+\sqrt{b}$ as a root, with $a,b \in \mathbb{Z} $, $b$ a prime number, so is $a-\sqrt{b}$."

REPLY [2 votes]: Another way, similar to Barry Cipra's method: 
Assume that $P$ is a polynomial with rational coefficients and let $a$ and $b$ be rationals such that $\sqrt{b}$ is irrational such that $P(a+\sqrt{b})=0$. 
By the binomial theorem, $P(a+\sqrt{b})+P(a-\sqrt{b})$ is rational. To see why, write $P(x)=\sum_{k=0}^nc_kx^k$, and expand $P(a \pm \sqrt{b})$; when you sum $P(a+\sqrt{b})$ and $P(a-\sqrt{b})$, the terms containing $\sqrt{b}$ cancel. 
By similar reasoning $P(a+\sqrt{b})-P(a-\sqrt{b})$ is some rational multiple of $\sqrt{b}$.
But $P(a+\sqrt{b})=0$, thus $P(a-\sqrt{b})$ is both rational and some rational multiple of $\sqrt{b}$. Hence $P(a-\sqrt{b})=0$.