What is the maximal possible number of roots on the interval (0,1) for a polynomial of degree 2022 with integer coefficients and with the leading coefficient equal to 1?
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Tags: algebra, polynomial
16.05.2022 10:18
iamworseatmaths wrote: What is the maximal possible number of roots on the interval (0,1) for a polynomial of degree 2022 with integer coefficients and with the leading coefficient equal to 1? Certainly not $2022$ (since then their product would be in $(0,1)$ and so the constant coefficient of polynomial would not be an integer) and at least $1011$, for example with polynomial $\prod_{n=1}^{1011}(x^2-(n+2)x+n)$
18.05.2022 10:20
iamworseatmaths wrote: What is the maximal possible number of roots on the interval (0,1) for a polynomial of degree 2022 with integer coefficients and with the leading coefficient equal to 1? Claim : the greatest number of roots in $(0,1)$ of a monic polynomaial $P(x)\in\mathbb Z[X]$ with degree $n$ is $n-1$ 1) this number is $<n$ If it was $n$, then the product of the $n$ roots in $(0,1)$ would be in $(0,1)$, which is impossible since this product is $(-1)^nP(0)\in\mathbb Z$ Q.E.D. 2) This number may indeed be $n-1$ Let $Q(x)=-\prod_{i=1}^{n-1}(i-nx)$ : this is a (non monic) polynomial of $\mathbb Z[X]$ with degree $n-1$ and $n-1$ roots in $(0,1)$ : $r_i=\frac in$where $i\in\{1,2,...,n-1\}$ Let "middles" $m_0=\frac{0+r_1}2$, $m_i=\frac{r_i+r_{i+1}}2$ for $i\in\{1,2,...,n-2\}$ and $m_{n-1}=\frac{r_{n-1}+1}2$ We have $Q(m_i)<0$ for all even $i$ and $Q(m_i)>0$ for all odd $i$ Let $u=\min_{i\in\{0,1,...,n-1\}}|Q(m_i)|$ and positive integer $k>\frac 1u$ Consider $P(x)=x^n+kQ(x)$. This is monic polynomial $\in\mathbb Z[X]$ Since $|kQ(m_i)|\ge ku>1$ and $m_i^n\in(0,1)$, we get that $P(m_i)$ has same sign that $Q(m_i)$ And so $P(m_i)<0$ for all even $i$ and $P(m_i)>0$ for all odd $i$ And so at least one real root for $P(x)$ in each interval $(m_i,m_{i+1})$ for any $i\in\{0,1,...,n-2\}$ And so at least $n-1$ real roots in $(0,1)$ And so exactly $n-1$ real roots in $(0,1)$ Q.E.D. Hence answer : $\boxed{2021}$