query: How are the factors obtained at *this point* in the solution of proving $2^{2^n}+ 1 \mid 2^{2^m} -1$
positive_passages.docid: 22093
positive_passages.text: How to show $a^{2^n}+1 \mid a^{2^m}-1$?
negative_passages.docid: 148330
negative_passages.text: Induction Proof for $F_{2n} = F^2_{n+1} - F^2_{n-1}$
negative_passages.docid: 44928
negative_passages.text: Show $\sum_n \frac{z^{2^n}}{1-z^{2^{n+1}}} = \frac{z}{1-z}$
negative_passages.docid: 119109
negative_passages.text: Integer solutions to $ n^2 + 1 = 2 \times 5^m$
negative_passages.docid: 44897
negative_passages.text: Proof: If $n=ab$ then $2^a-1 \mid 2^n-1$
negative_passages.docid: 142715
negative_passages.text: Manipulating expression $(\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}})$ in inductive step.
negative_passages.docid: 78632
negative_passages.text: Summation equation for $2^{x-1}$
negative_passages.docid: 51858
negative_passages.text: Showing $\gcd(2^m-1,2^n+1)=1$
negative_passages.docid: 127293
negative_passages.text: Prove that $(a-b) \mid (a^n-b^n)$
negative_passages.docid: 66558
negative_passages.text: Find the value of $a$ and $b$ if $x+1$ and $x−2$ are factors of $ax^3−4x^2+bx−12$
negative_passages.docid: 119118
negative_passages.text: Prove $n\mid \phi(2^n-1)$
negative_passages.docid: 10742
negative_passages.text: Powers of prime factors of a positive integer $n$ in "Mathematica"?
negative_passages.docid: 108550
negative_passages.text: Elements of $\mathbb Q(2^{1/3})$
negative_passages.docid: 113230
negative_passages.text: How can we factor $ q^4 + q^2 + 1$
negative_passages.docid: 138683
negative_passages.text: Proving $\prod_{i=2}^{i=n} \left(1-\frac{1}{i^2}\right) = \frac{n+1}{2n}$ by induction
negative_passages.docid: 40262
negative_passages.text: Factoring $-3x^2 - 2xy + 8y^2 + 11x - 8y - 6$
negative_passages.docid: 69136
negative_passages.text: factorization of a^n+1?