About real roots of A polynomial with real coefficients that has three consecutive equal

Question

How to prove : A polynomial with real coefficients that has three consecutive equal (non-zero) coefficients of successive powers of the variable cannot have all its roots real.

I know how to prove it using Viète's relations in specific cases. For example, in the case where the polynomial starts or ends with 3 terms with equal coefficients. I get lost in the general case, and I think Descartes' rule will be helpful.

Addition

According to Descartes, we have the following

lemma:

• If $P(X)=\sum_{k=0}^n a_k X^k$ and There exist integers $i$, $j$, and $k$ such that $j < i$, $i + 1 < k$, $a_j \cdot a_k \neq 0$, and $a_i = a_{i+1} = 0$ then P cannot have all its roots as real numbers.

Let's assume that our polynomial is written as $P(X) = X^k + X^{k+1} + X^{k+2} + (\text{other terms})$. Now, consider the polynomial $Q(X) = (X-1)P(X) = XP(X) - P(X)$. When we expand the calculations, We observe that Q satisfies the assumptions of the lemma. Then we deduce that $Q$ cannot have all its roots as real numbers. Consequently, $P$ cannot have all its roots as real numbers either.

Answer 1

A proof can be given using Rolle's theorem. Wlog we can assume the three consecutive coefficients are all $1$ as the division of a polynomial by a nonzero constant doesn't affect the zeroes.

Let the coefficients be in position $k, k+1, k+2, k \ge 0$ and assume $P(x)$ of degree $n+k, n \ge 2$ has all $n+k$ real roots so by Rolle theorem $P^{(k)}(x)$ of degree $n$ has $n$ real roots (inductively there are $n+k-1$ roots of the derivative, one in-between each pair of roots of $P$ etc).

But $P^{(k)}(x)=k!+(k+1)!x+(k+2)!x^2/2+...a_n x^n, n \ge 2$.

Now clearly $$Q(x)=x^nP^{(k)}(1/x)=k!x^n+(k+1)!x^{n-1}+(k+2)!x^{n-2}/2+..+a_n$$ also has all real roots, so again by Rolle $$Q^{(n-2)}(x)=k!n!x^2/2+(k+1)!(n-1)!x+(k+2)!(n-2)!/2$$ has all real roots.

However the discriminant of $Q^{(n-2)}(x)$ is $$k!(k+1)!(n-2)!(n-1)!((k+1)(n-1)-(k+2)n)<0$$ which is a contradiction, so we are done!