Polynomials with integer coefficients – find minimal $m$

Question

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 $m$?

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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.

Answer 1

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.

Answer 2

Our goal is to find $p,q$ in $\mathbb{Z}[x]$ such that

$$p(x)f(x) + q(x)g(x) = 32760$$

where

\begin{align*} f(x) &= (x+4)(x+5)(x+9)\\[6pt] g(x) &= (x-4)(x-5)(x-9)\\[6pt] \end{align*}

To find $p,q$, apply the Extended Euclidean Algorithm . . .

Step $1\,$: Dividing $f(x)$ by $g(x)$ yields

$$f(x) = (1)g(x) + (36x^2+360)$$

hence

$$36x^2+360 = f(x) - g(x)$$

Step $2\,$: Dividing $g(x)$ by $36x^2+360$ yields

$$g(x)= \left(\frac{1}{36}x - \frac{1}{2}\right)(36x^2+360)+ (91x)$$

hence

\begin{align*} 91x &= g(x) - \left(\frac{1}{36}x - \frac{1}{2}\right)(36x^2+360)\\[6pt] &= g(x) - \left(\frac{1}{36}x - \frac{1}{2}\right)(f(x) - g(x))\\[6pt] &= \left( -\frac{1}{36}x + \frac{1}{2} \right) f(x) + \left( \frac{1}{36}x + \frac{1}{2} \right) g(x) \end{align*}

Step $3\,$: Dividing $36x^2+360$ by $91x$ yields

$$36x^2+360 = \left(\frac{36}{91}x\right)(91x) + (360)$$

hence

\begin{align*} 360 &= (36x^2+360) - \left(\frac{36}{91}x\right)(91x)\\[6pt] &= (f(x) - g(x)) - \left(\frac{36}{91}x\right) \left(\left( -\frac{1}{36}x + \frac{1}{2} \right) f(x) + \left( \frac{1}{36}x + \frac{1}{2} \right) g(x) \right)\\[6pt] &= \left(\frac{1}{91}x^2 - \frac{18}{91}x + 1\right) f(x) + \left(-\frac{1}{91}x^2 - \frac{18}{91}x - 1\right) g(x) \end{align*}

Clearing denominators,

\begin{align*} & \left(\frac{1}{91}x^2 - \frac{18}{91}x + 1\right) f(x) + \left(-\frac{1}{91}x^2 - \frac{18}{91}x - 1\right) g(x) = 360\\[10pt] \implies\; &(x^2-18x+91)f(x) + (-x^2-18x-91)g(x) = 32760 \end{align*}

Thus, if

\begin{align*} p(x) &= x^2 - 18x + 91\\[6pt] q(x) &= -x^2 - 18x - 91 \end{align*}

then

$$p(x)f(x) + q(x)g(x) = 32760$$