This is an exercise that I solved before many semesters, and I just remember I solved in a very complicated(or stupid) way, today when I need to use this exercise as a tool I wonder if there are some brief proofs
Thanks for anyone’s help!
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1Hint: Show that $\gcd(x^a -1, x^b -1) = x^{\gcd(a, b)}-1$. – Calvin Lin May 02 '21 at 01:40
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The "only if" is the harder direction. Let $j$ be the smallest nonegative integer s.t. $d|(n-j)$. Assume $j$ is positive. Then $j<d$ and [assuming the "if" direction] $$(x^d-1)|(x^{n-j}-1)x^j$$ $$ = x^n-1-(x^j-1).$$ So the remainder of $x^n-1$ divided by $x^d-1$ is nonzero, precisely a polynomial of degree precisely $j$, namely $x^j-1$.
To see the "if" direction let $y=x^d$. Then $x^d-1=y-1$ and $x^n-1=y^{\frac{n}{d}}-1$ where $\frac{n}{d}$ is a positive integer.
Mike
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Please strive not to add more dupe answers to dupes of FAQs. See site policy here. – Bill Dubuque May 02 '21 at 02:18