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Let $f(x)$ be a polynomial with integer coefficients. If $f(x)$ is divisible by $x^2+1$ and $f(x)+1$ by $x^3+x^2+1$, what is $f(x)$?

My guess is that the only answer is $f(x)=-x^4-x^3-x-1$, but how can I prove it?

中山優燈
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  • How did you come to that result? – graydad Jun 12 '15 at 15:18
  • @graydad just guessing, I would say... – 中山優燈 Jun 12 '15 at 15:28
  • Sometimes there are well-founded reasons to make guesses at how a solution will look. I was just wondering if this was one of those times. Without such reasoning it is probably better to invoke rigor to find a solution, than to just (somewhat) randomly guess. I wrote up an answer below that outlines a way to find an $f$ that satisfies your conditions. – graydad Jun 12 '15 at 15:32

3 Answers3

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In fact, there are infinitely many solutions. If $f(x)=-x^4-x^3-x-1$ and $g(x)$ is any polynomial, then

$$f(x)+g(x)(x^2+1)(x^3+x^2+1)$$

is a solution.

ajotatxe
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This is a straightforward application of CRT = Chinese Remainder Theorem.

By hypothesis $\ (x^2\!+1)g = f = -1 + (x^3\!+x^2\!+1)h\ $ for some polynomials $\,g,h,\,$ so

${\rm mod}\ x^2\!+\!1\!:\ x^2\equiv -1\,\Rightarrow\,1\equiv (x^3\!+x^2\!+1)h\equiv -xh\,\Rightarrow\, h \equiv -1/x\equiv x^2/x\equiv \color{#c00}x$

Therefore $\ h = \color{#c00}x + (x^2\!+1)h',\ $ for some polynomial $\,h'$

$\! \begin{align}{\rm hence}\ \ f\, =&\, -1 + (x^3\!+x^2\!+1)(\color{#c00}x+(x^2\!+1)h')\\ =&\ \ x^4\!+x^3\!+x-1 + (x^3\!+x^2\!+1)(x^2\!+1) h'\end{align}$

Bill Dubuque
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One way you could express this idea is through functions $g,h$ where $$f(x) = g(x)(x^2+1) \\ f(x)+1 = h(x)(x^3+x^2+1)$$ This means that $$g(x)(x^2+1)+1 =h(x)(x^3+x^2+1)$$ or that $$h(x) = \frac{g(x)(x^2+1)+1}{x^3+x^2+1}$$ So you could begin by choosing any polynomial $g$ you want so long as it forces $h$ to be a polynomial. Then $f$ will be a polynomial.

graydad
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