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 4 of 5: Even case: peel off m_k = n + 2^k − 2

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1+\frac{2^k-1}{2t}=\left(1+\frac{2^{k-1}-1}{t}\right)\left(1+\frac{1}{2t+2^k-2}\right)

Detailed analysis

If $n=2t$ , direct computation gives $\dfrac{2t+2^k-1}{2t}=\dfrac{2t+2^k-2}{2t}\cdot\dfrac{2t+2^k-1}{2t+2^k-2}$ , which is exactly the displayed product once each fraction is written as $1+(\ldots)$ . Applying the induction hypothesis to $t$ and $k-1$ produces $m_1,\ldots,m_{k-1}$ for the first factor, and $m_k=2t+2^k-2=n+2^k-2$ is a positive integer supplying the second.