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.

  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.