Polynomials with integer coefficients – find minimal $m$ Let $m$ be the minimal positive integer such that
$$(x+4)(x+5)(x+9)p(x) - (x-4)(x-5)(x-9)q(x) = m$$ 
where $p(x)$ and $q(x)$ are polynomials with integer coefficients. What is the value of $m$?

My work :
By Bezout's identity, we have, $\exists p(x), q(x) \in \mathbb{Z}[x]$ such that
$(x+4)(x+5)(x+9)p(x) - (x-4)(x-5)(x-9)q(x) = gcd((x+4)(x+5)(x+9), (x-4)(x-5)(x-9))$
Consider the order pair, $((p(x),-q(x))$, by Euclid's Algorithm,
$gcd((x+4)(x+5)(x+9),  (x-4)(x-5)(x-9))$ 
$ = gcd(x^3-18x^2+101x-180, x^3+18x^2+101x+180)$ 
$ = gcd(x^3-18x^2+101x-180, 36x^2 + 360)$
Please suggest how to proceed.

LEMMA: If $f(x),g(x)\in \mathbb Z[x]$ and a prime number $p$ divides every co-efficient of the product $f(x)g(x)$ then $p$ divides every co-efficient of $f(x)$ or of $g(x).$
From this we  can show that if $n\in \mathbb N$ divides every co-efficient of $f(x)$ and $n$ is co-prime to some (any) co-efficient of $g(x)$ then $gcd (f(x),g(x))=gcd (f(x)/n,g(x)).$
From your last line therefore $$gcd (x^3-18x^2+101x-180, 36x^2+360)=gcd (x^3-18x^2+101x-180,x^2+10)=$$ $$=gcd (x^3-18x^2+101x-180-x(x^2+10),x^2+10)=gcd (-18x^2+91x-180,x^2+10)=$$ $$=gcd(-18x^2+91x-180+18(x^2+10),x^2+10)=$$ $$=gcd (91x,x^2+10)=gcd(x,x^2+10)=$$ $$=gcd(x,x^2+10-x(x))=gcd(x,10)=gcd(x,1)=1.$$ Note that the disappearance of the factors "$36$" and the "$10$" are due to the lemma.
PROOF OF LEMMA: Let $f(x)=\sum_i f_ix^i$ and $g(x)=\sum_j g_jx^j$ and $f(x)g(x)=\sum_kh_kx^k.$ Suppose by contradiction  that the prime $p$ divides every $h_k$ but not every $f_i$ and not every $g_j.$   Let $i_1$ be the least $i$ such that $p\not | \;f_i$ and let $j_1$ be the least $j$ such that $p\not |\; g_j.$
We have $h_{(i_1+j_1)}=\sum_{i+j=i_1+j_1}f_ig_j.$ In this sum :(i) When $k<i_1$ we have $p|f_i,$ and (ii) When $k>i_1$ we have $j<j_1$ so $p|g_j.$ Therefore , modulo $p,$ we have $$0\equiv h_{(i_1+j_1)}\equiv f_{i_1}g_{j_1}\not  \equiv 0,$$ a contradiction.
