Problem 1

Prove that for any pair of positive integers

k

and

n

, there exist

k

positive integers

m_1,m_2,\ldots,m_k

(not necessarily different) such that

1+\frac{2^k-1}{n}=\left(1+\frac{1}{m_1}\right)\left(1+\frac{1}{m_2}\right)\cdots\left(1+\frac{1}{m_k}\right).

Step 2 of 5: Inductive step: split n by parity

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n=2t-1 \text{ (odd)}\quad\text{or}\quad n=2t \text{ (even)}, \quad t\in\mathbb{Z}_{>0}

Detailed analysis

Fix $k\ge2$ and assume the identity is already known for $k-1$ and every positive integer. Given $n$ , write $n=2t-1$ if $n$ is odd or $n=2t$ if $n$ is even, for a positive integer $t$ ; the two cases are treated separately below.