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I need to find (if they exists) $h(x)$ and $k(x)$ in $\mathbb{Q}[x]$ such that

$20 = h(x) (x^2 +1) + k(x) (x^3 -3)$

my manual does not provide any method to compute the unknown functions, I was suggested to use the Euclidean extended algorithm but since it is not treated in the book I'd prefer not using it.

gioretikto
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    Let $h(x)=ax^2+bx+c$, $k(x)=dx+e$, multiply everything out, solve for the unknowns $a,b,c,d,e$ (but Euclid is much easier). – Gerry Myerson Sep 12 '20 at 09:35
  • How have you chosen the definition of h(x) and k(x)? – gioretikto Sep 12 '20 at 09:38
  • I used a good statement of Bezout. – Gerry Myerson Sep 12 '20 at 09:39
  • Note that if you know the Euclidean algorithm, the extended version is obvious: it only relies an the fact that, not only the g.c.d. satisfies a Bézout's identity, but all remainders in the process do. Other than that, you can start from the last equality in the succession of Euclidean divisions, and go backwards. – Bernard Sep 12 '20 at 09:52
  • To expand on my comment: if $h,k$ is a solution, and $p$ is any polynomial, then $h-(x^3-3)p,k+(x^2+1)p$ is also a solution, and then you can choose $p$ so that $h-(x^3-3)p$ has degree two (by taking $p$ to be the quotient on division of $h$ by $x^3-3$), and then automatically (by considering the degrees) $k+(x^2+1)p$ will have degree one. Try it! – Gerry Myerson Sep 12 '20 at 13:13
  • You have several answers now, gioretikto. Any thoughts about any of them? – Gerry Myerson Sep 14 '20 at 10:40
  • It's not polite to ask people for help and then, when they help you, to run away and hide. Please engage with us, gioretikto. – Gerry Myerson Sep 16 '20 at 02:34

3 Answers3

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$\!\!\bmod\,\overbrace{ x^2+1}^{\color{#c00}{\textstyle \!\!x^2\!\equiv\! -1}\!\!\!}\!:\,\ {-}20 \equiv (\overbrace{3-x\color{#c00}{x^2}}^{\!\!\!\!\!\textstyle 3+x})\,k$ $\iff k \equiv\dfrac{-20^{\phantom{|^|}}}{x+3} \equiv -20\left[ \dfrac{x\, -\, 3}{\smash{\underbrace{\color{#c00}{x^2}-9}_{\large\!\!\! -10}}}\right] \equiv 2(x-3)$

Remark $ $ If mod arithmetic is unfamiliar we can do the same by evaluating the hypothesized Bezout equation $\, 20 = k(x) (x^2-3) + h(x) (x^2+1)\,$ at $\,x = i^{\phantom{|^|}}\!\!,\,$ yielding

$$20 = k(i)(-i-3)\iff k(i) = \dfrac{\!\!-20}{i+3} = 2(i-3)\ \ \rm as\ above\quad $$

By theory we know there is a solution with $\,\deg k < \deg(x^2+2),\,$ so $\,k = ax+b,\,$ so $\, 2i-6 = k(i) = ai+b \iff (2-a)i = b+6\iff a=2,\, b=-6,\,$ by $\,i\not\in \Bbb Q$

Bill Dubuque
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  • Remark $ $ Since $\Bbb Q[x]/(x^2+1)\cong \Bbb Q[i],$ we are essentially calculating with $,x,$ as $,i,, $ so the calculation can be viewed as Gaussian integer arithmetic if we replace $,x,$ by $,i\ \ $ – Bill Dubuque Sep 12 '20 at 22:18
  • Alternatively apply the Extended Euclidean Algorithm as here or here (they explain the general idea behind the method in Bernard's answer). – Bill Dubuque Sep 12 '20 at 22:25
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Use the extended Euclidean algorithm (same as for $\mathbf Z$):

\begin{array}{r| c c l} r(x)&u(x)&v(x) & q(x) \\ \hline x^3-3 & 0 & 1 \\ x^2+1& 1 & 0 & x \\ \hline -x-3 & -x & 1 & -x+3 \\ 10 &1+3x-x^2 & x-3 \\ \hline \end{array} Therefore, one obtains $$ 10 =(1+3x-x^2)(x^2+1) +(x -3)(x^3-3)$$ whence $\quad 20=2(1+3x-x^2)(x^2+1)+2 (x-3)(x^3-3)$.

Bernard
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Just to show that it can be done: $$20=(x^2+1)(ax^2+bx+c)+(x^3-3)(dx+e)$$ $$=(a+d)x^4+(b+e)x^3+(a+c)x^2+(b-3d)x+c-3e$$ so $$a+d=0, b+e=0, a+c=0, b-3d=0,c-3e=20$$ so $$b=3d,e=-b=-3d,c=-a=d,20=c-3e=d+9d=10d$$ hence $$d=2,b=6,e=-6,c=2,a=-2$$ and we have $$h(x)=-2x^2+6x+2,k(x)=2x-6$$

Gerry Myerson
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