A polynomial 


$$f(x) = x^n + a\_{n-1} x^{n-1} + \cdots + a\_1 x + a\_0$$ 


where $a\_{n-1}, a\_0$ are known. How to determine whether polynomial $f$ has integer roots? Can we give a bound of $\max\{a\_i \mid i=1,\dots,n-2\}$ and $\min\{a\_i \mid i=1,\dots,n-2\}$ decide by $n,a\_{n-1}$, and $a\_0$ such that the polynomial roots are integers.



How to determine whether a polynomial has integer roots?
I'm not sure if I understand just what you're asking, but perhaps it is the following:



> 
> Suppose we know $a\_0$ and $a\_{n-1}$, and we know that $f(x)$ has only integer roots. Can we use this information to meaningfully bound the coefficients?
> 
> 
> 


If that's what you're asking, then the answer is no. Consider for example the polynomial
$$
g(x) = x (x^2 - c^2)^{k},
$$
where $c \in \mathbb{Z}$. Then this has $a\_0 = a\_{n-1} = 0$. But even for fixed $n$, the other coefficients are unbounded if you let $c$ be large.




---


This part added in response to Timothy Chow's comment:
------------------------------------------------------


Timothy Chow made the suggestion that allowing $a\_{0} = a\_{n-1} = 0$ feels like cheating, and on reflection, I agree!


**If we allow $a\_0 = 0$, then we can't get any bound on the other coefficients**


Namely, just consider any monic degree $n-2$ polynomial of the form $p(x) = x^{n-2} + b\_{n-4} x^{n-4} + b\_{n-5} x^{n-5} + \cdots + b\_0$. Then look at the polynomial $x (x+a) p(x) = x^n + a x^{n-1} + a\_{n-2} x^{n-2} + \cdots + a\_1 x + 0$. This will have integer roots, and the coefficient $a\_{n-1}$ is free to be any integer you like. Yet the other coefficients can be wild. For concreteness, consider this
$$
h(x) = x (x+a)(x^2 - c^2)^k,
$$
which, for fixed $a$ and $k$, will have unbounded coefficients if you pick $c$ large enough [e.g., the sum of these coefficients is $h(1)$ which is unbounded].


**If we disallow $a\_0 = 0$, then we do get a bound**


However! If we disallow the case $a\_0 = 0$, then *yes* you can get a bound on the remaining coefficients. Namely, if $f(x)$ factors as $f(x) = \prod\_{i} (x +r\_i)$, then we know $a\_0 = \prod\_{i} r\_i$. If none of the roots is $0$ (i.e., if $a\_0 \neq 0$), then each root is at least $1$ in absolute value, and thus $|a\_0|$ is an upper bound on every product of the form $\prod\_{i \in S} |r\_i|$. Therefore, we have


$$
|a\_{n-k}| = \left| \sum\_{|S| = k} \prod\_{i \in S} r\_i \right| \leq \sum\_{|S| = k} \prod\_{i \in S} |r\_i| \leq \sum\_{|S| = k} |a\_0| = {n \choose k} |a\_0|.
$$
This bound is tight of course by the polynomials $(x+1)^n$ and $(x-1)^n$. (This reminds me of Newton's inequalities, which probably can provide an alternate proof.)


