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

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.

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.

• Why is that thingy about Descartes true? $x^2$ has two consecutive zero coefficients and two real zeroes? Commented Nov 17, 2023 at 13:42
• Also any $x^2Q(x)$ with real roots $Q$ works though indeed I suspect those are the only counterexamples to your claim Commented Nov 17, 2023 at 13:49
• @Conrad I apologize; I hadn't specified the lemma, which is a consequence of Descartes. I have added the relevant lemma
Commented Nov 17, 2023 at 14:53
• yes that looks good now - note that again all follows by Rolle here too, though in a sense Rolle (and calculus) is not really needed, just the fact that the convex hull of the roots moves downward with derivatives for complex polynomials which is a very simple algebraic property seen by considering $f'/f$ (which again can be thought of as a formal expression, no calculus needed), so the roots of the derivative are included in the convex hull of the roots of the polynomial; in particular the convex hull of the reals is the reals etc Commented Nov 17, 2023 at 16:24
• Thank you very much for your support.
Commented Nov 17, 2023 at 16:52

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!

• Thank you very much for this beautiful proof. In the meantime, I believe I have succeeded in a proof using Descartes. If anyone is interested, I can add it to my initial message to see if it's correct