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It is well known that $2^{ab}-1=(2^a-1)(1+2^a+...+2^{(b-1)a})=(2^b-1)(1+2^b+...+2^{(a-1)b})$. Assuming $gcd(2^a-1,2^b-1)=1$, we see $2^b-1|(1+2^a+...+2^{(b-1)a})$. My question is simply how to factor out the factor $2^b-1$ from this expression.

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    $\begingroup$ Try a few concrete cases first, like $a=2,b=3$, and $a=2,b=5$, and $a=3,b=4$, and $a=3,b=5$. See if you can spot some immediate pattern. If you swap $2$ with $x$, then this wikipedia article might help you identify the factors which appear. $\endgroup$
    – Arthur
    Aug 18 '18 at 4:34
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Since the primitive part of a term in a sequence is defined as the term divided by the algebraic part, and because here the algebraic part is $(2^a-1)(2^b-1)$, we know the expression we are trying to find is the primitive part of $2^{ab}-1$. This is calculable via $\Phi_{ab}(2)$, using roots of unity as shown in the link provided in the comment.

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