The question
Let $a \in (0, \infty)$ and the set
$$M=\{a+x+\sqrt{a-x+x^2}|x\in N*\}$$
Show that:
a)if $a=1$, then the set $M$ contains exactly $2$ rational numbers
b) if the set $M$ contains at least $2$ rational numbers, then the number $a$ is rational
The idea
a) if $a=1$ then we have to solve $1+x+\sqrt{1-x+x^2}=y$, where $y$ is an rational number
$\sqrt{x^2-x+1}=y-x-1$ and because $y-x-1$ is an rational number we get that $\sqrt{x^2-x+1}$ should also be a rational number $=> x^2-x+1$ is a perfect square
$x^2-x+1=k^2$
Maybe I can demonstrate it by using a quadric formula…I don't know.
b) For point b we have to show that if $x^2-x+a=k^2$ has at least $2$ rational solutions, then $a$ is rational…
I don't know what to do forward. I hope one of you can help me. Thank you!
Best Answer
For the second part (since you've already solved the first):
Let the 2 values of $x$ be $m$ and $n$. Then: $$(a+m + \sqrt{m^2-m+a}) - (a+n + \sqrt{n^2-n+a}) \in \Bbb Q \iff \sqrt{m^2-m+a} - \sqrt{n^2-n+a}\in \Bbb Q$$ Note that $\sqrt{m^2-m+a} - \sqrt{n^2-n+a}$ is non-zero since $x^2-x$ is increasing over the naturals. Then: $$\frac{n-m}{\sqrt{m^2-m+a} + \sqrt{n^2-n+a}} \in \Bbb Q \iff \sqrt{m^2-m+a} + \sqrt{n^2-n+a} \in \Bbb Q$$ Adding, $\sqrt{m^2-m+a} \in \Bbb Q$, so this forces $a$ to be rational.