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 1 of 5: Base case k = 1

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In plain words

Peeling off one factor at a time will reduce $k$ to $k-1$ , so induction on $k$ is natural.

1+\frac{2^1-1}{n}=1+\frac{1}{n}

Detailed analysis

For $k=1$ the claimed identity reads $1+\frac{1}{n}=1+\frac{1}{n}$ , which holds trivially by taking $m_1=n$ . This is the base case of an induction on $k$ that must hold for every positive integer $n$ .