Sum of roots of an irreducible polynomial in Q[x].

Question

Set $f(x)\in \mathbb{Q}[x]$ is irreducible, and $\deg f(x)=2n+1$. Prove that sum of any two different roots of $f(x)$ can't be a rational number.

Answer 1

Lemma 1: For a polynomial P(x) not in $Q[X]$ there exists atmost one non constant monic polynomial with minimial degree G(x) such that $P(x) G(x) \in \mathbb{Q}[X]$ , and all other polynomials are divisible by $G(x)$

Proof :- Say there exists another polynomial $K(x)$ such that $P(x)K(x) \in \mathbb{Q}[X]$.Consider the remainder formed by we divide $P(x)K(x)$ by $P(x)G(x)$ , that should also be a multiple of $P(x)$[very easy to proof left as an excercise to reader ] as well say $P(x)T(x)$ and due to division algorithm $P(x)T(x) \in \mathbb{Q}[X]$.And degree of $T(x)$ is less than $G(x)$ contradiction.

Now let $f(x)=(x+\alpha_1)(x+\alpha_2).... (x+\alpha_n)$ and $\alpha_1+\alpha_2 \in \mathbb{Q}$ [Note:- I am assuming f(x) is monic here, if it's not proof can be changed easily just write a multiplied by C at front ]

Clearly as degree is odd $n \geq 3$ .

And $(x+\alpha_3)(x+\alpha_4).... (x+\alpha_n)$ is irrecuible using lemma 1 this implies that it's the unique polynomial with minimal degree. .

let $\alpha_1+\alpha_2=s$

then consider $T(x)=(x+s-\alpha_1)(x+s-\alpha_2)(x+s-\alpha_3).... (x+s-\alpha_n)$ can be easily shown to be in $Q[X]$

However $T(x)=(x+\alpha_1)(x+\alpha_2)(x+s-\alpha_3).... (x+s-\alpha_n)$

Using lemma 1 that implies

$(x+s-\alpha_3)..... (x+s-\alpha_n)=(x+\alpha_3)(x+\alpha_4).... (x+\alpha_n)$

This implies for each $\alpha_j$ it can be paired $\alpha_i$ such that their sum is $s$.However if $i=j$ then $\alpha_i$ is rational contradicting irreducibility. So they are disticnt however that implies $f(x)$ has even degree.

Answer 2

Only thing I can think of is that the polynomial $g(x) = (x-a)(x-b) = x^2 - (a+b)x + ab$, with $a,b$ being roots of $f$ divides $f$ (in $\mathbb{C}[x]$) and can only be in $\mathbb{Q}[x]$ if $a+b$ is rational.

My guess would be that somehow the fact that the degree is odd should imply $ab$ to be rational as well.