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I have to determine the lowest degree of $P$ given by the following system : $\left\{ \begin{array}{l} P \equiv 2X \ \mod[X^2 -2X +1] \\ P \equiv 3X \ \mod[X^2 -4X+4] \end{array} \right.$

First, I say that $\exists \ Q_1, Q_2 \in K[X]$ such as :

$P(X)=(X^2-2X+1)Q_1(X)+2X$ and $P(X)=(X^2-4X+4)Q_2(X)+3X$

Then I obtain : $(X^2-2X+1)Q_1(X)-(X^2-4X+4)Q_2(X)=-X$

Then I proved by euclidian divison that: $\gcd(X^2-2X+1,X^2-4X+4)=1$

So according to Bezout theorem, $\exists \ U, V \in K[X]$ such as :

$(X^2-2X+1)U(X)+(X^2-4X+4)V(X)=1$

I've tried to find a particular solution for $U,V$ and to solve it like a diophantine equation but I think it does not work for polynoms. By intuition I said that $U(X)=aX+b$ and $V(X)=cX+d$ and found $U(X)=-2X+5$, $V(X)=2X-1$

Now I meet difficulties to find $P$

Thanks in advance.

Maman
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  • The two polynomials you're reducing w.r.t. are perfect squares. Also, try the Chinese Remainder Theorem. – kodlu May 21 '15 at 12:16
  • If you have run the Euclidean Algorithm to find the gcd, you can calculate "backwards" to find $U(X)$ and $V(X)$, in the same way as you may have done it earlier with integers. – André Nicolas May 21 '15 at 12:20
  • @AndréNicolas I tried this method and I found $U(X)=-2X+5$ and $V(X)=2X-1$ which was my particular solution. I got it by identification and by taking $U,V$ in the form $aX+b$. – Maman May 21 '15 at 12:50
  • If you can find $U_1(X), V_1(X)$ that give you $1/4$, you can multiply them by $4$ to obtain $1$ as a linear combination. – André Nicolas May 21 '15 at 12:50
  • @AndréNicolas I did what you said and found these polynoms but the final $P$ is not correct. – Maman May 21 '15 at 12:57
  • You were asking about $U$ and $V$. They can be a part of a strategy for finding $P$ using the same idea as in the proof of the Chinese Remainder Theorem. Or else you can "guess" at the shape as suggested by Michael, and calculate. Your method based on subtraction need not work, having the difference be $-X$ does not guarantee that we have congruence to $2X$ and $3X$. – André Nicolas May 21 '15 at 13:05

2 Answers2

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Let $\,a=x\!-\!1, b=x\!-\!2.\,$ Squaring a Bezout Identity (BI) for $\,\color{#c00}{a,b}\,$ yields a BI for $\,\color{#0a0}{a^2,b^2}$

$$1 = \color{#c00}{a-b}\overset{\rm square}\Rightarrow 1 = a^2\!+b^2\!-2ab(\color{#c00}{a\!-\!b})\ =\, (1\!-\!2b)\,\color{#0a0}{a^2}\!+(1\!+\!2a)\,\color{#0a0}{b^2}\qquad $$

Finally, from the BI for $\,\color{#0a0}{a^2,b^2}$ we can read off the CRT solution in the usual way, $ $ i.e.

$\quad$ if $\ u \color{#0a0}{a^2}\! + v\color{#0a0}{b^2} = 1\ $ then $\ P = d ua^2\! + cvb^2\,$ $\Rightarrow$ $\, P\equiv c\pmod{\!a^2},\ P\equiv d\pmod{\!b^2}$

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Bill Dubuque
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  • You're right with your method I found the same $U$ and $V$, but now I don't know how to determine $P$ – Maman May 21 '15 at 15:20
  • @Maman I added a remark on that. – Bill Dubuque May 21 '15 at 15:45
  • $c$ and $d$ are polynoms ? – Maman May 21 '15 at 16:17
  • Then we have to find $c$ and $d$ ? – Maman May 21 '15 at 16:58
  • @Maman Yes, $,a,b,c,d, =, (x!-!1)^2,,(x!-!2)^2,, 2x,, 3x,$ in your case, and $,u,v,=, 1!-!2b,, 1!+!2a,=, 5!-!2x,, 2x!-!1.\ \ $ – Bill Dubuque May 21 '15 at 16:58
  • You are given $,c,d,,$ i.e. they are the RHS of the CRT system, $,P\equiv c\pmod{a^2},\ P\equiv d\pmod{b^2}\ \ $ – Bill Dubuque May 21 '15 at 16:59
  • I found $P=-2x^4+29x^3-76x^2+58x$ it does not work – Maman May 21 '15 at 17:11
  • @Maman Right, that is not correct. Check your arithmetic. – Bill Dubuque May 21 '15 at 17:19
  • @Maman The cofactors of $,(x-1)^2,$ and $,(x-2)^2,$ are wrong. Swap them to fix it, i.e. $,dua^2 = (3x)(5-2x)(x-1)^2$ – Bill Dubuque May 21 '15 at 17:44
  • Thanks it works !!! But now how do we know that this is the lowest degree ? – Maman May 21 '15 at 17:54
  • Reduce it mod $,F = (x-1)^2(x-2)^2,$ to get a solution $P$ of degree $< \deg F.,$ Such a solution is unique by CRT. Or, directly: if $,P'$ were another solution of deg $\le 3,$ then $,P\equiv P',$ mod $,(x!-!1)^2$ and mod $(x!-!2)^2$ so also mod their lcm = product $= F$. Then $,F\mid P-P',,$ so $,P-P' = 0$ since it has smaller degree than $F$ (this is the proof of the uniqueness (mod $F)$ of the CRT solution). – Bill Dubuque May 21 '15 at 17:59
  • Ok so we apply CRT to solve this kind of problems. Interesting. – Maman May 21 '15 at 18:05
  • @Maman Yes, just as for integers, polynomials (over a field) enjoy a Division with Remainder algorithm, so they enjoy a Euclidean algorithm, Bezout identity, etc. A conveneient way to do the Extended Euclidean algorithm (to get $U,V)$ is in this answer. – Bill Dubuque May 21 '15 at 18:08
  • Ok thanks, is it possible to find another $P$ with $\deg(P)=4$ ? – Maman May 21 '15 at 18:20
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    @Maman If $P$ is a solution then $P'$ is a solution $\iff P'\equiv P\pmod F.,$ Therefore the complete set of solutions is $\ P + Q F,$ for all polynomials $Q.,$ This implies that there exists a solution of degree $,d\iff d\ge 3.\ \ $ – Bill Dubuque May 21 '15 at 18:31
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You could say that $P(1)=2$, $P'(1)=2$, $P(2)=6$, $P'(2)=3$. This is four equations in the coefficients of $P(x)$. Let $P(x)=Ax^3+Bx^2+Cx+D$, and you have four equations in four coefficients. For example, $P(1)=A+B+C+D=2$.

Empy2
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