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For what $m$ and $n$ is the polynomial $2X^{19}+X^{13}+mX^{11}+X^8+2X^6+nX^2+2$ divisible by $X^4+X^3+X^2+X+1$.

I tried to find the real solutions for g but couldn't

lioness99a
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Lola
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2 Answers2

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HINT:

$$x^5-1=(x-1)(x^4+x^3+x^2+x+1)=0$$

$$\implies x^{19}=x^4,x^{13}=x^3$$

lab bhattacharjee
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Note that $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$

\begin{align} &\;2x^{19}+x^{13}+mx^{11}+x^8+2x^6+nx^2+2\\ =&\;2x^4(x^5-1+1)^3+(x^3+mx)(x^5-1+1)^2+(x^3+2x)(x^5-1+1)+nx^2+2\\ =&\;2x^4[(x^5-1)^3+3(x^5-1)^2+3(x^5-1)+1]\\ &\quad+(x^3+mx)[(x^5-1)^2+(x^5-1)+1]+(x^3+2x)[(x^5-1)+1]+nx^2+2\\ =&\;P(x)(x^5-1)+2x^4+x^3+mx+x^3+2x+nx^2+2 \end{align}

for some polynomial $P(x)$.

$2x^{19}+x^{13}+mx^{11}+x^8+2x^6+nx^2+2$ is divisible by $x^4+x^3+x^2+x+1$ if and only if $2x^4+x^3+mx+x^3+2x+nx^2+2$ is divisible by $x^4+x^3+x^2+x+1$.

\begin{align} &\;2x^4+x^3+mx+x^3+2x+nx^2+2\\ =&\;2x^4+2x^3+nx^2+(m+2)x+2 \end{align}

It is obvious that $m=0$ and $n=2$.

CY Aries
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