[Anonymous314]

Let $P(x),Q(x)$ be integer polynomials and $m \in \mathbb{N}$ such that
$$(x+4)(x+5)(x+9)P(x)-(x-4)(x-5)(x-9)Q(x)=m.$$
Find $\min(m)$ that satisfies the equation.

Answer: 32,760
Solution:
Let $$f(x)=(x+4)(x+5)(x+9)P(x)-(x-4)(x-5)(x-9)Q(x)=m.$$
We have
$$m=f(4)=8\cdot 9\cdot 13P(4)$$
$$m=f(5)=9\cdot 10 \cdot 14 P(5)$$
$$m=f(9)=13 \cdot 14 \cdot 18 P(9).$$
From $P(x)$ is a integer polynomial, if one of $P(4),P(5),P(9)$ is $0$ we get a contradiction from $m \in \mathbb{N}$ thus,
$$LCM[8\cdot 9\cdot 13,9\cdot 10 \cdot 14,13 \cdot 14 \cdot 18]|m$$
or
$$32,760|m.$$
For $x \in \mathbb{Z}$ we will find the maximum value of $$GCD[(x+4)(x+5)(x+9),(x-4)(x-5)(x-9)].$$
$$GCD[(x+4)(x+5)(x+9),(x-4)(x-5)(x-9)]=GCD[x^3+18x^2+101x+180,x^3-18x^2+101x-180]=GCD[36x^2+360,x^3-18x^2+101x-180]$$
$$GCD[36x^2+360,x^3-18x^2+101x-180]|36 \cdot GCD[x^2+10,x^3-18x^2+101x-180]=36 \cdot GCD[x^2+10,18x^2-91x+180=36 \cdot GCD[x^2+10,-91x]$$
$$36 \cdot GCD[x^2+10,-91x]| 36 \cdot 91 \cdot GCD[x^2+10,x]=36 \cdot 91 \cdot GCD[10,x]$$
$$36 \cdot 91 \cdot GCD[10,x]| 36 \cdot 91 \cdot 10=32,760$$
and we have $$GCD[(x+4)(x+5)(x+9),(x-4)(x-5)(x-9)]|32,760.$$

From the fact that the equation ,For $k,a,b \in \mathbb{Z}$ there are infinitely many pairs $(x,y) \in \mathbb{Z} \times \mathbb{Z}$ such that
$$ax+by=kGCD(a,b)$$ and the minimum positive integer $c$ that satisfies the equation $$ax+by=c$$ is $GCD(a,b)$.

Thus $$\min(m)=32,760$$ as desired.