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For a polynomial $f(X)\in \mathbb C[X]$, and $a\in \mathbb C$, let $f^{-1} (a):=\{\mu \in \mathbb C : f(\mu)=a\}$.

Now let $f(X), g(X) \in \mathbb C[X]$ be non-constant polynomials such that $f^{-1}(0)=g^{-1}(0)$ and $f^{-1}(1)=g^{-1}(1)$, then is it true that $f=g$ ?

Let $f^{-1}(0)=\{\mu_1,...,\mu_k\}; f^{-1}(1)=\{\gamma_1,...,\gamma_l\}$.

Then $f(X)=c_1\prod_{i=1}^k(X-\mu_i)^{n_i}, f(X)-1=c_1'\prod_{i=1}^l(X-\gamma_i)^{n_i'}$

$g(X)=c_2\prod_{i=1}^k(X-\mu_i)^{m_i}; g(X)-1=c_2'\prod_{i=1}^l(X-\gamma_i)^{m_i'}$

But I don't know how to proceed further.

Please help

user521337
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  • A polynomial is determined by its zeroes, counted by multiplicity, along with one more constant factor. Knowing $f^{-1}(0)$ determines the zeros, but not their multiplicity. – Jair Taylor Dec 09 '18 at 01:46
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    @JairTaylor but different multiplicities will tend to put the solutions of $f(x)=1$ at different points. – Empy2 Dec 09 '18 at 02:05
  • @Empy2 Ah, you are right. I see now based on the answer below. – Jair Taylor Dec 09 '18 at 17:48

1 Answers1

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We may assume without loss of generality that $\deg f\geq \deg g$. Let $d=\deg f$, let $A=f^{-1}(0)=g^{-1}(0)$, and let $B=f^{-1}(1)=g^{-1}(1)$. Note that $|A|=d-m$ where $m$ is the number of roots that $f'$ has in $A$ (counting multiplicity). Similarly, $|B|=d-n$ where $n$ is the number of roots that $f'$ has in $B$. Thus $$|A|+|B|=2d-(m+n)\geq 2d-(d-1)=d+1$$ (since $f'$ has $\deg f'=d-1$ roots in total).

But now notice that every point of $A\cup B$ is a root of $f-g$. Since $f-g$ has degree at most $d$ and $|A\cup B|=|A|+|B|\geq d+1$, this implies that $f-g=0$ and so $f=g$.

Eric Wofsey
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