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Given $d,B>0$ the number of polynomials in $\mathbb Z[x]$ of degree $d$ and coefficient size at most $B$ have at least one integer roots should be $B^{O(d)}f(d)$ at some function $f$ (from Random Diophantine polynomials: Percent solvable?). So the probability that a uniformly random polynomial in $\mathbb Z[x]$ of degree $d$ and coefficient size at most $B$ has at least one integer root is $\frac{B^{O(d)}f(d)}{(2B+1)^{d+1}}\asymp\frac{f(d)}{2B+1}$ which is close to $0$.

  1. Then what is the average number of integer roots for polynomials in $\mathbb Z[x]$ of degree $d$ and coefficient size at most $B$?

I think it should be $<1$.

However I am not sure of the exact parameterizations.

It might be $\frac1{B^{O(d)}}$ on average.

  1. What is the average number of integer roots for polynomials in $\mathbb Q[x]$ of degree $d$ and coefficients with numerator of size at most $B$ and denominator of size at most $C$?
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  • $\begingroup$ When you say "$f(B)^{d - 1}$", do you mean that it should grow with $O(B^{d - 1})$? Because neither is true, for example, in the $d = 1$ case, where the growth is logarithmic (an integer root means the coefficient of $x$ divides the coefficient of $1$, so you get approximately the harmonic series sum). $\endgroup$ – user44191 Jan 22 at 10:26
  • $\begingroup$ @user44191 $bx-a$ has integer root if $b|a$ and $b=1$ itself implies $a$ has $O(B)$ possibilities and so we have at least $B$ degree $1$ polynomials. $\endgroup$ – T.... Jan 22 at 10:28
  • $\begingroup$ Apologies; I should have said there are $B \text{ln} (B)$ possibilities (approximately). $\endgroup$ – user44191 Jan 22 at 10:36
  • $\begingroup$ Maybe go at it from the other direction – how many polynomials of degree $d$ and coefficients bounded by $B$ have, say, 17 as a root? Then take a sum, as 17 varies, to get the total number of integer roots, and then the average number. $\endgroup$ – Gerry Myerson Jan 22 at 11:01
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    $\begingroup$ Over $\mathbb R$, the answer is zero with any reasonable notion of "average". With probability 1 the coefficients will be algebraically independent, so they won't have a single integer, rational nor even algebraic root. $\endgroup$ – Wojowu Jan 22 at 12:24
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$$\mathbb{E}[\text{#|roots of P|}]=\mathbb{E}(\sum_{k\in \mathbb{Z}}1_{k \text{ is a root of P}})=\sum_{k\in\mathbb{Z}}\mathbb{P}(P(k)=0)$$ For $k=0$, $\mathbb{P}(P(0)=0)=\frac{1}{(2B+1)}$.

For $k\neq 0$, because the coefficients are independent (if $B$ and $d$ large) $P(k)$ should behave like a Gaussian if $d$ is large of variance $\sigma^2=\sum_{j=0}^d \big(k^{2j}\times\frac{1}{(2B+1)}\sum_{i=-B}^Bi^2\big)\approx (d+1)\times \frac{B^2}{3}$ if $|k|=1$ and $\frac{k^{2d+2} -1}{k^2-1} \times \frac{B^2}{3}$ for $|k|>1$. So $$ \mathbb{P}(P(k)=0)\approx \begin{cases} \frac{1}{B\sqrt{2\pi(d+1)/3}} \text{ for }k=1 \\ \frac{1}{B\sqrt{2 \pi\frac{k^{2d+2} -1}{k^2-1}/3}} \text{for |k|>1}\end{cases}$$ Note that for large $d$, the probability for $|k|>1$ is very small and then $$\mathbb{E}[\text{#|roots of P|}]\approx \mathbb{P}(P(0)=0)+\mathbb{P}(P(1)=0)+\mathbb{P}(P(-1)=0)\approx\frac{1}{(2B+1)}(1+2\times \frac{\sqrt{6}}{\pi d}) $$(same result as in Random Diophantine polynomials: Percent solvable?)

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  • $\begingroup$ Why should they behave Gaussian? $\endgroup$ – T.... Jan 22 at 12:02
  • $\begingroup$ Would similar result hold in $\mathbb R[x]$? $\endgroup$ – T.... Jan 22 at 12:12

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