Let $a$ be a positive integer, but not a perfect square; $r$ is a real root of the equation $x^3-2ax+1=0$. Prove that $ r+\sqrt{a}$ is an irrational number.
Problem
Source: China Zhongshan ,13 Aug 2014
Tags: algebra, polynomial, number theory proposed, number theory
mavropnevma
13.08.2014 13:13
Assume $r+\sqrt{a} = s\in \mathbb{Q}$; then $r = s-\sqrt{a}$ verifies $r^3 -2ar +1 = 0$, thus $(s^3 +sa+1) - (3s^2-a)\sqrt{a} = 0$. Therefore $s^3 +sa+1 = 3s^2-a = 0$, whence $4s^3 + 1 = 0$, absurd.
Naysh
15.08.2014 10:24
Suppose to the contrary that $c=r+\sqrt{a}$ is a rational number.
Since $a$ is not a perfect square, the minimal polynomial of $\sqrt{a}$ is $x^2-a$. However, $\sqrt{a}$ is also a root of $(c-x)^3-2a(c-x)+1=0.$
Thus, we must have $(x^2-a) | x^3-3cx^2+(c^2-2a)x-(c^3-2ac+1)$.
The quotient of the two polynomials must be of the form $x+b$. Comparing constant terms, we see that $b=\dfrac{c^3+1}{a}-2c$.
Then we have $x^3-3cx^2+(c^2-2a)x-(c^3-2ac+1)=(x^2-a)\left(x+\dfrac{c^3+1}{a}-2c\right)$. Comparing the coefficent of $x$ on both sides, we see that $c^2-2a=-a\Longrightarrow a=c^2$, which contradicts the assumption that $a$ is not a perfect square.
Thus $c$ cannot be rational, and the desired assertion is proved.
SCP
22.08.2014 21:46
mavropnevma wrote: Therefore $s^3 +sa+1 = 3s^2-a = 0$, whence $4s^3 + 1 = 0$, absurd. I couldn't see why this would mean $4s^3 + 1 = 0$. But different finish: First let $s= \frac{p}{q}$ where $p,q$ are integers and $gcd(p,q)=1$. Now looking to the coefficients of $q$ : we have $q|1$. Hence $s \in \mathbb Z$. Now $a= s^2-2s-2 - \frac{1}{s-1}$ is an integer, so $s=0$ of $2$. But both values imply that $a$ is negative, contradiction.
mathocean97
23.08.2014 03:19
SCP wrote: mavropnevma wrote: Therefore $s^3 +sa+1 = 3s^2-a = 0$, whence $4s^3 + 1 = 0$, absurd. I couldn't see why this would mean $4s^3 + 1 = 0$. Well, since $3s^2-a = 0$, $a = 3s^2$. Substituting this into $s^3+as+1 = 0$, $s^3+(3s^2) \cdot s+1 = 0 \implies 4s^3 +1 = 0$.