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 5 of 5: Conclude the induction

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1+\frac{2^k-1}{n}=\prod_{i=1}^{k}\left(1+\frac{1}{m_i}\right)

Detailed analysis

Every positive integer $n$ is either odd or even, so one of the two cases above always applies. Together with the base case $k=1$ , this completes the induction on $k$ and proves the identity for every pair of positive integers $k,n$ .