1

Is there a program, small script, or easy method I can use to find the polynomial P(x) such that:

$P(x) = x+1$ $\pmod {x^2+1}$

$P(x) = x+2$ $\pmod {x^2+2}$

$P(x) = x+3$ $\pmod {x^2+3}$

$P(x) = x+4$ $\pmod {x^2+4}$

.........

This is easy to do with integers, but I can't get across this with polynomials. Thanks for help!

J. Linne
  • 3,022

4 Answers4

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Hint Since $k=-X^2 \pmod{X^2+k}$, you have $$P(X) \equiv X-X^2 \pmod{X^2+1} \\ P(X) \equiv X-X^2 \pmod{X^2+2} \\ P(X) \equiv X-X^2 \pmod{X^2+3} \\ P(X) \equiv X-X^2 \pmod{X^2+4} \\$$

N. S.
  • 132,525
1

Use the constructive proof of the Chinese remainder theorem.

The solution of the first two ones is $x-x^2$, obtained using Bézout's identity for $x^2+1$ and $x^2+2$.

It so happens that $x-x^2$ works for all other congruences: $x-x^2=(-1)(x^2+n)+(x+n)$.

lhf
  • 216,483
1

It didn't take too long to compute the following table of moduli.

\begin{array}{r|cccc} & x^2 + 1 & x^2 + 2 & x^2 + 3 & x^2 + 4 \\ \hline \dfrac 16(x^2+2)(x^2+3)(x^2+4) & 1 & 0 & 0 & 0 \\ -\dfrac 12(x^2+1)(x^2+3)(x^2+4) & 0 & 1 & 0 & 0 \\ \dfrac 12(x^2+1)(x^2+2)(x^2+4) & 0 & 0 & 1 & 0 \\ -\dfrac 16(x^2+1)(x^2+2)(x^2+3) & 0 & 0 & 0 & 1 \\ \hline \end{array}

It follows that

\begin{align}P(x) = & \dfrac 16(x^2+2)(x^2+3)(x^2+4)(x+1)- \\ & \dfrac 12(x^2+1)(x^2+3)(x^2+4)(x+2)+ \\ & \dfrac 12(x^2+1)(x^2+2)(x^2+4)(x+3)- \\ & \dfrac 16(x^2+1)(x^2+2)(x^2+3)(x+4) \end{align}

With the help of Wolfram alpha, this expands to $f(x) = x-x^2$

Steven Alexis Gregory
  • 26,769
1

Note that if $\ \color{#c00}{a_i - b_i} = c\ $ is constant (wrt $i)$ then below reduces to CCRT = Constant-case CRT $$\ \{\,f \equiv \color{#c00}{a_i}\!\!\!\pmod{\color{#c00}{b_i}}\,\}_{i=1}^k\! \iff \{\,f \equiv c\!\!\pmod{b_i}\,\}_{i=1}^k\iff f\equiv c\!\!\pmod{{\rm lcm}\{b_i\}}$$

since $\ a_i \equiv b_i+c\equiv c\pmod{b_i}.\ $ This common optimization applies to your CRT system.

Remark $ $ Generally the solution will have form $\ f \equiv c \equiv a_i + d_i b_i $ where the $d_i$ may depend on $i$ and need not be "small". But in human-constructed problems (exercises, contests) it if often worth testing small or constant $d_i,\,$ e.g. the obvious choice $\,d_i = -1\,$ in OP to eliminate dependence on $i$ in the constant terms by subtracting the modulus from the residue $\ x+i-(x^2+i) \equiv x-x^2.$

Bill Dubuque
  • 272,048