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Let $m$ be the minimal positive integer such that

$(x+4)(x+5)(x+9)p(x) - (x-4)(x-5)(x-9)q(x)=m$.

$p(x)$ 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.

  • 3
    Use the Euclidean Algorithm to actually find polynomials $p(x),q(x)$ with integer coefficients such that $$(x+4)(x+5)(x+9)p(x) - (x-4)(x-5)(x-9)q(x)=32760$$ – 2017-02-04
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    @ quasi, will you please explain in more detail on how to use Euclidean Algorithm to find the polynomials. – 2017-02-04
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    Lookup "The Extended Euclidean Algorithm". It's a little tedious to carry out by hand. If you have access to Maple, you can use the gcdex command to get p(x),q(x) (although the coefficients are rational, so you'll need to clear denominators). – 2017-02-04
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    If I had more time, I could show the exact steps, but I have no time right now. Searching for "polynomial extended euclidean algorithm" should get you on the right track. – 2017-02-04

2 Answers 2

1

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

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

  • 0
    Please look at the 2nd line of step 2, $g(x) = \left(\frac{1}{36}x - \frac{1}{2}\right)(36x^2+360)+ (91x)$, since $p, q$ must be in $\mathbb{Z}[x]$ but I think $ \left(\frac{1}{36}x - \frac{1}{2}\right) $ may not be integer for all $x$. Please explain on this point. – 2017-02-06
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    @carat: It's only the final line that counts -- the one that yields an identity of the form $p(x)f(x) + q(x)g(x) = 32760$. – 2017-02-06
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    I'm not sure what you are confused about, but perhaps the last few lines of the current edit will make it clearer. – 2017-02-06
  • 0
    That's just long division of polynomials in $\mathbb{Q}[x]$ – 2017-02-06
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    Though $p, q$ are in $\mathbb{Z}[x]$, but long division of polynomials must be done in $\mathbb{Q}[x]$, right ? – 2017-02-06
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    Yes, all the long divisions were done in $\mathbb{Q}[x]$. – 2017-02-06
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    All the equations are identities in $\mathbb{Q}[x]$, but the last identity involves only integer polynomials, so it's also an identity in $\mathbb{Z}[x]$. – 2017-02-06
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    The goal was to find polynomials $p,q$ in $\mathbb{Z}[x]$ such that $$p(x)f(x) + q(x)g(x) = 32760$$. If you expand and simplify, you can check that the polynomials $p,q$ given by \begin{align*} p(x) &= x^2 - 18x + 91\\ q(x) &= -x^2 - 18x - 91 \end{align*} satisfy the requirements. – 2017-02-06
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    I understand the rest. Thank you both. – 2017-02-06
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 $$gcd (x^3-18x^2+101x-180, 36x^2+360)=gcd (x^3-18x^2+101x-180,x^2+10)=$$ $$=gcd (x^3-18x^2+101x-180-x(x^2+10),x^2+10)=gcd (-18x^2+91x-180,x^2+10)=$$ $$=gcd(-18x^+91x-180+18(x^2+10),x^2+10)=$$ $$=gcd (91x,x^2+10)=gcd(x,x^2+10)=$$ $$=gcd(x,x^2+10-x(x))=gcd(x,10)=gcd(x,1)=1.$$ 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 $ki_1$ we have $j

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    The lemma is also useful in proving a result of Gauss, that if $f\in \mathbb Z[x]$ is irreducible in $\mathbb Z[x]$ then $f$ is irreducible in $\mathbb Q[x].$ – 2017-02-06