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I need to find $f(x)\in \mathbb{Q}[x]$, such that:

  1. $f(x)\equiv 1 \pmod{(x-1)^2}$
  2. $f(x)\equiv x \pmod{x^2}$
  3. $\deg(f(x))<4$

So, what I understand so far is that:

  1. $(x-1)^2\mid f(x)-1$
  2. $x^2\mid f(x)-x$
    (I hope I'm right at those)

But I'm stuck here, I don't know how to continue...
Can you help me please?

Thank you!

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5 Answers 5

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Let the polynomial be $f(x)=ax^3+bx^2+cx+d$. The second condition is easier to apply, because it implies $c=1$ and $d=0$.

Now let's apply the first condition. You know that $f(1)-1=0$, so $$ a+b+1-1=0 $$ On the other hand, $1$ must also be a root of the derivative of $g(x)=f(x)-1$. Hence $$ 3a+2b+1=0 $$ Thus $a=-1$ and $b=1$.

The polynomial is $f(x)=-x^3+x^2+x$.

Without derivatives: consider $g(x)=f(x)-1=ax^3+bx^2+x-1$. It should be divisible by $x-1$, so $g(1)=a+b+1-1=0$. Hence $b=-a$. Now we see that $$ g(x)=ax^3-ax^2+x-1=ax^2(x-1)+(x-1)=(x-1)(ax^2+1) $$ This should be divisible by $(x-1)^2$, yielding $a=-1$.

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  • $\begingroup$ Worth emphasis: we can avoid the need to solve systems of equations (or use derivatives) by instead using an operational form of CRT, which uses only trivial mental arithmetic - see my answer. $\endgroup$ Jul 23, 2019 at 15:13
  • $\begingroup$ I nee dto do it without derivative ... :( $\endgroup$
    – CS1
    Jul 24, 2019 at 19:45
  • $\begingroup$ @CS1 Just do division, then. $\endgroup$
    – egreg
    Jul 24, 2019 at 20:09
  • $\begingroup$ @egreg division of what? Thank you! $\endgroup$
    – CS1
    Jul 24, 2019 at 20:16
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    $\begingroup$ @CS1 I added the argument. $\endgroup$
    – egreg
    Jul 24, 2019 at 20:43
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$f(x)-1=g(x-1)^2$ with $g$ that is a polynomial of degree $1$, so $g=ax+b$ but you have also that

$f(x)-x=(cx+d)x^2$

so

$(ax+b)(x-1)^2+1=(cx+d)x^2+x$

Then $a=c$, $-2a+b=d$, $a-2b=1$, $b+1=0$

so

$b=-1$, $a=-1$, $d=1$, $c=-1$.

Then

$f(x)=-(x+1)(x-1)^2+1$

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$-(1+x)(x-1)^{2}+1$ is such a polynomial.

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  • $\begingroup$ This is correct, but will likely mystify beginners. It can be derived very simply using the modular distributive law - see my answer. $\endgroup$ Jul 23, 2019 at 15:22
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Using $\ ag\,\bmod\, bg\, =\, g\,\left[\,a\bmod b\,\right]\:$ to factor out $\: g = (x\!-\!1)^{\large 2}$ $\,\ \begin{align}f\!-\!1\!\!\!\pmod{\!x^{\large 2} g} &\,=\, g\,\left[\dfrac{\!\color{#c00}f\!-\!1\,}{(x\!-\!1)^{\large 2}}\bmod {x^{\large 2}}\right]^{\phantom{|^|}}\!\!,\, \ {\rm so}\ \ \color{#c00}{f\equiv x}\!\!\!\pmod{\!x^2}\ \ \rm yields\\ &\!\!\!=\ {-}g\,\left[\,\color{#0a0}{x\!+\!1}\,\right]\ \ {\rm by}\ \ \dfrac{\color{#c00}x\!-\!1\ }{(x\!-\!1)^{\large 2}} \equiv \dfrac{1}{x\!-\!1} \equiv \dfrac{1\!-\!\color{#90f}{x^{\large 2}}}{x\!-\!1} \equiv \color{#0a0}{-(x\!+\!1)}\!\!\pmod{\!\color{#90f}{x^{\large 2}}} \end{align}$

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Consider the polynomial $f(x)=(Ax+B)(x-1)^2+1$. It is evident that $f$ satisfies the conditions 1 and 3. It remains to find $A$ and $B$ such that $$f(x)=(Ax+B)(x^2-2x+1)+1\equiv (A-2B)x+B+1\equiv x \pmod{x^2}$$ that is $B=-1$ and $A=-1$. Hence $$f(x)=-(x+1)(x-1)^2+1=-x^3+x^2+x.$$

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    $\begingroup$ Dear downvoter, a comment will help me to improve my answer. $\endgroup$
    – Robert Z
    Jul 23, 2019 at 18:05
  • $\begingroup$ @CS1 I edited my answer, is it clear now? $\endgroup$
    – Robert Z
    Jul 25, 2019 at 4:41

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