# How does one show that a given polynomial is bijective?

Let $P:\mathbb{R} \rightarrow \mathbb{R}$ be a given polynomial function. We wish to determine the bijectivity of $P$. The first thing that comes to mind is showing that the derivative does not change sign. However, in my view, this is not trivial for high-degree polynomials.

This problem can be solved easily for specific types of polynomials. Let $P(x)=mx+b$ with $m≠0$, a polynomial of degree one. It is injective since $m(x+c) + b = mx + b \Longrightarrow c=0$. It is clearly surjective since the inverse $P^{-1}(x)=m^{-1}(x-b)$ is defined for all real $x$. Other observations we can make are that if we have a polynomial of the form $x^n + c$ for $n$ odd, this is bijective, that a polynomial of even degree is never bijective, and a degree three polynomial $ax^3 + bx^2 + cx + d$ with $a≠0$ is bijective iff $3ac \geq b^2$. The proofs of these assertions are not difficult.

It gets difficult, however, when we have nontrivial odd degree polynomials of $n \geq 5$. This is where I am unsure how to proceed.

• One (overkill) option would be to compute the Sturm sequence and show that the derivative has no real roots. – Michael Burr Feb 14 '16 at 22:58
• @MichaelBurr: why overkill? – Rob Arthan Feb 14 '16 at 23:00
• @RobArthan It's a lot to compute a Sturm sequence and if the OP was interested in a specific class of polynomial, there may be a more specialized answer. But, a Sturm sequence will always work. – Michael Burr Feb 14 '16 at 23:06

Let $Q$ be a polynomial. Compute $D=\gcd(Q,Q')$. Then the roots of $D$ are precisely the multiple roots of $Q$ and are of multiplicity one less. Therefore $Q/D$ is a polynomial with the same roots as $Q$, but all roots are simple. Finally $R=\gcd(Q/D,D)$ has precisely the multiple roots of $Q$ as roots and in fact as simple roots. Let's call $R$ the simplification of $Q$.
We need that $\deg P$ is odd and that every critical point (i.e., root of $P'$) is a root of $P'$ of even multiplicity. Compute a sequence of polynomials $F_0,F_1,\ldots$ as follows: Let $F_0=P'$. For each $n$, let $R_n$ be the simplification of $F_n$ and $F_{n+1}=F_n/R^2$. We can stop as soon as $R_n$ is constant. Then any root of $F_n$ is a root of odd multiplicity of $P'$. Hence what remains is to show that $F_n$ has no real roots. This can be done with Sturm sequences.