MathLabs

Problem 1

Prove that for any pair of positive integers kk and nn, there exist kk positive integers m1,m2,…,mkm_1,m_2,\ldots,m_k (not necessarily different) such that 1+2k−1n=(1+1m1)(1+1m2)⋯(1+1mk).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 3 of 5: Odd case: peel off m_k = n
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1+2k−12t−1=(1+12t−1)(1+2k−1−1t)1+\frac{2^k-1}{2t-1}=\left(1+\frac{1}{2t-1}\right)\left(1+\frac{2^{k-1}-1}{t}\right)
Detailed analysis

If n=2t−1n=2t-1, then (2t−1)+(2k−1)=2t(1+2k−1−1t)(2t-1)+(2^k-1)=2t\left(1+\frac{2^{k-1}-1}{t}\right); dividing by 2t−12t-1 gives the displayed factorization. By the induction hypothesis applied to tt and k−1k-1, the second factor equals (1+1m1)⋯(1+1mk−1)\left(1+\frac{1}{m_1}\right)\cdots\left(1+\frac{1}{m_{k-1}}\right) for suitable positive integers, and taking mk=2t−1=nm_k=2t-1=n completes this case.