# Integral roots of univariate polynomials

# Taha Azzaoui - 2018.04.20

### Brief Introduction

Lecturing before the International Congress of Mathematicians in 1900, the great David Hilbert presented the
world with a list of 23 unsolved mathematical conundrums, which in his expert opinion, were worthy of pursuit. The
tenth among them seeks the derivation of a method which, given any finite polynomial with integer coefficients, can
decide if said polynomial has integral roots (roots belonging to the set of integers). As an example, *f*(*x*, *y*, *z*) = *x* + *y*^{2} + *z*^{3}
has an integral root at ( − 31, 2, 3) but *g*(*x*) = *x*^{2} − 2 has no integral roots (since $\sqrt{2} \notin \mathbb{Z}$).

To quote Hilbert, problem ten roughly translates from his native German to the following: “Given a Diophantine equation with any number of unknown quantities and with rational integral numerical coefficients: To devise a process according to which it can be determined in a finite number of operations whether the equation is solvable in rational integers.” For now, think of a Diophantine equation as essentially a polynomial. We will define it more rigorously soon enough.

Such an algorithm was though to exist (as implied by Hilbert’s question) until Yuri Vladimirovich Matiyasevich et al. eloquently proved otherwise. For the sake of intuition, before we get into Matiyasevich’s insight and the computability of Diophantine sets, let’s consider the familiar case of finite polynomials with integer coefficients over only one variable, and show that there does exist an algorithm for determining their integral roots.

### A plausible solution

Since the integers are countable, our first approach in devising an algorithm would be to evaluate the
polynomial at ± *i* for all *i* ∈ ℕ and simply halt once we find a root. However, consider the case of a polynomial
with no integral roots. It’s clear that the algorithm will run forever, unless we can come up with some tight bound
on the roots, outside of which we are certain no roots exist.

### Bounding the roots

Consider the finite polynomial with integer coefficients of order n…

*f*(*x*) = *c*_{0} + *c*_{1}*x* + *c*_{2}*x*^{2} + *c*_{3}*x*^{3} + ... + *c*_{n}*x*^{n}

Suppose *r* is a root of the polynomial …

0 = *c*_{0} + *c*_{1}*r* + *c*_{2}*r*^{2} + *c*_{3}*r*^{3} + ... + *c*_{n}*r*^{n}

We can rearrange the terms and yield …

− (*c*_{0} + *c*_{1}*r* + *c*_{2}*r*^{2} + *c*_{3}*r*^{3} + ... + *c*_{n − 1}*r*^{n − 1}) = *c*_{n}*r*^{n}

We can then take the absolute value of both sides …

|*c*_{0} + *c*_{1}*r* + *c*_{2}*r*^{2} + *c*_{3}*r*^{3} + ... + *c*_{n − 1}*r*^{n − 1}∥ = ∥*c*_{n}*r*^{n}|

We then apply the triangle inequality …

|*c*_{0}∥ + ∥*c*_{1}*r*∥ + ∥*c*_{2}*r*^{2}∥ + ∥*c*_{3}*r*^{3}∥ + ... + |*c*_{n − 1}*r*^{n − 1}| ≥ |*c*_{n}*r*^{n}|

Denote the coefficient with the largest magnitude by *c*_{m}.

The inequality still holds if we substitute it in place of the remaining coefficients…

|*c*_{m}|(1 + |*r*| + |*r*^{2}| + |*r*^{3}| + ... + |*r*^{n − 1}|) ≥ |*c*_{n}*r*^{n}|

Note: $(1 + |r| + |r^2| + |r^3| + ... + |r^{n-1}|) \leq \sum_{i = 0}^{n} |r^{n - 1}\| =
n|r^{n - 1}|$

Therefore …

$$|c_m| |r^{n - 1}|n \geq |c_nr^n| \implies \frac{|c_m|}{|c_n|}n \geq |r|$$

In other words, the integral roots of the polynomial must lie within the interval $[-\frac{c_m}{c_n}n,\frac{c_m}{c_n}n]$. Where *c*_{m} is the coefficient with the largest magnitude, *c*_{n} is the coefficient of the highest order term, and *n* is the order of the polynomial.

### Exhaustive but finite

Now that we have a tight bound on the roots of the polynomial, we can devise an algorithm whose running time is finite. We proceed the same as before, except we now evaluate the polynomial at every integer within the range $[-\frac{c_m}{c_n}n, \frac{c_m}{c_n}n]$. We simply return true if the function evaluates to 0 and false if the last value in the set is reached and no such root was found. Below is python script that implements this exhaustive search.

```
"""
Reads in a set of n coefficients {c_0, c_1, ..., c_n} as
command line arguments such that f(x) = sum(c_i * x^i)
as a command line arguments and attempts to find
the integral roots of the polynomial they represent.
"""
import sys
import math
def f(coefs, x):
"""
Evaluates f(x) given a
set of coefficients
and an input x.
"""
res = 0
for i in range(len(coefs)):
res += coefs[i] * x**i
return res
# input: {c_0, c_1, ..., c_n} s.t. f(x) = sum(c_i * x^i)
coefs = [int(i) for i in sys.argv[1:]]
# Order of the polynomial
order = len(coefs) - 1
# Coefficient with the largest magnitude
c0 = max([abs(i) for i in coefs])
# Cofficient of the highest order term
c1 = coefs[order]
# +- bound on possible values for the roots of f(x)
bound = math.ceil(order*c0/c1)
# set of all possible values for the roots of f(x)
values = list(range(-1 * bound, bound))
root = None
for val in values:
print("f({}) = {}".format(val, f(coefs,val)))
if f(coefs, val) == 0:
print("Found integral root! x = {}".format(val))
root = val
if root == None:
print("No integral roots found!")
```