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Find the remainder when $x^{10}+1$ is divided by $(x^2+1)(x^2+x+1)$ I have done it until the the divisor is of second degree. But here the degree of the remainder is $4$ This means the remainder will be of the form. : $ax^3+bx^2+cx+d$ which makes it difficult to solve. Is it possible to solve $4$ variants with $4$ equations?

lulu
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John757
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    $$\dfrac{x^{10}+1}{(x^2-1)(x^2+x+1)}=(x-1)(x^5-x^3+x^2+x-1)+\dfrac{{x^3+x}}{(x^2-1)(x^2+x+1)}$$ – Piquito Jul 20 '18 at 15:09
  • Your method should have been to use Division of Polynomials, which should have been taught to you in high school. No simultaneous equations. – Lubin Jul 23 '18 at 18:02

3 Answers3

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Using Proof of $a^n+b^n$ divisible by $a+b$ when $n$ is odd,

$$x^{10}+1\equiv0\pmod{x^2+1}$$

Again $x^3-1=(x-1)(x^2+x+1)$

$$x^{10}+1= x(x^9-1)+x+1\equiv x+1\pmod{x^2+x+1}$$

Now apply CRT

lab bhattacharjee
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$$\dfrac{x^{10}+1}{(x^2-1)(x^2+x+1)}=(x-1)(x^5-x^3+x^2+x-1)+\dfrac{{x^3+x}}{(x^2-1)(x^2+x+1)}$$

John757
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Note that: $$\frac{x^{10}+1}{(x^2+1)(x^2+x+1)}=\frac{(x^2+1)(x^8-x^6+x^4-x^2+1)}{(\color{green}{x^2+1})(x^2+x+1)}$$ Using long division: $$ \require{enclose} \begin{array}{r} \color{blue}{x^6-x^5-x^4+2x^3-2x+1} \\[-3pt] x^2+x+1 \enclose{longdiv}{x^8-x^6+x^4-x^2+1} \\[-3pt] \underline{x^8+x^7+x^6}\phantom{2} \\[-3pt] -x^7-2x^6+x^4-x^2+1 \\[-3pt] \underline{-x^7-x^6-x^5} \\[-3pt] -x^6+x^5+x^4-x^2+1 \\[-3pt] \underline{-x^6-x^5-x^4} \\[-3pt] 2x^5+2x^4-x^2+1 \\[-3pt] \underline{2x^5+2x^4+2x^3} \\[-3pt] -2x^3-x^2+1 \\[-3pt] \underline{-2x^3-2x^2-2x} \\[-3pt] x^2+2x+1 \\[-3pt] \underline{x^2+x+1} \\[-3pt] \color{red}x \\[-3pt] \end{array} $$ Hence: $$x^{10}+1=(x^2+1)(x^2+x+1)(\color{blue}{x^6-x^5-x^4+2x^3-2x+1})+\color{red}x(\color{green}{x^2+1}).$$

farruhota
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