Prove that ${x}^{4}+{x}^{3}+{x}^{2}+x+1\mid {x}^{4n}+{x}^{3n}+{x}^{2n}+{x}^{n}+1$

kadejoset
2022-07-18
Answered

Prove that ${x}^{4}+{x}^{3}+{x}^{2}+x+1\mid {x}^{4n}+{x}^{3n}+{x}^{2n}+{x}^{n}+1$

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Eve Good

Answered 2022-07-19
Author has **18** answers

${X}^{4}+{X}^{3}+{X}^{2}+X+1=\frac{{X}^{5}-1}{X-1}$

${X}^{4n}+{X}^{3n}+{X}^{2n}+{X}^{n}+1=\frac{{X}^{5n}-1}{{X}^{n}-1}$

Now, use the fact that ${X}^{5}-1|{X}^{5n}-1$ and that for n not divisible by 5 we have

$gcd({X}^{5}-1,{X}^{n}-1)={X}^{gcd(5,n)}-1=X-1$

${X}^{4n}+{X}^{3n}+{X}^{2n}+{X}^{n}+1=\frac{{X}^{5n}-1}{{X}^{n}-1}$

Now, use the fact that ${X}^{5}-1|{X}^{5n}-1$ and that for n not divisible by 5 we have

$gcd({X}^{5}-1,{X}^{n}-1)={X}^{gcd(5,n)}-1=X-1$

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