sqing wrote: Let $P(x)$ be a polynomial of degree $n\geq 2$ with rational coefficients such that $P(x) $ has $ n$ pairwise different reel roots forming an arithmetic progression .Prove that among the roots of $P(x) $ there are two that are also the roots of some polynomial of degree $2$ with rational coefficients . Let the $n$ roots be $\{r_k=a+kb\}_{k=1}^n$ $q_1=\sum r_k=na+\frac{n(n+1)}2b$ is a rational number $q_2=\sum r_k^2=na^2+n(n+1)ab+\frac{n(n+1)(2n+1)}6b^2$ is too a rational number So $\frac{q_2}n-\left(\frac{q_1}n\right)^2=\frac{n^2-1}{12}b^2$ is a rational number which means $b^2$ rational Then $r_1+r_n=2a+(n+1)b=\frac{2q_1}n$ is a rational number And $r_1r_n=a^2+(n+1)ab+nb^2=\frac{q_2}n-\frac{2n^2-3n+1}6b^2$ is a rational number And so $(r_1,r_n)$ is an example of requested pair (in fact $(r_k,r_{n+1-k})$ would be too)

roots $a,a+d,a+2d,...,a+kd$ $ a + (a+d) + (a+2d) + ... + (a+kd)$ is ratsional number.Hence $2a+kd$ ratsional number. $ \sum_{i<j} (a+di)(a+dj)$ is ratsional number. $a^2 \cdot \frac {k(k+1)}{2} + ad \cdot \frac{k^2(k+1)}{2} + d^2 \cdot \frac{(1+2+...+k)^2-1^2-2^2-...-k^2}{2}$ is ratsional number. $a^2+kad+d^2\cdot \frac{k^2}{4}$ is ratsional number.hence $d^2$ is ratsional number. $a^2+kad=a(a+kd)$ and $a+(a+kd)$ is ratsional number. there is a quadratic equation with rational coefficients whose roots are $a$ and $a+kd$.