Problem 2

Find all integers

n

satisfying

n\ge2

and

\frac{\sigma(n)}{p(n)-1}=n,

in which

\sigma(n)

denotes the sum of all positive divisors of

n

, and

p(n)

denotes the largest prime divisor of

n

Step 1 of 3: Upper bound on p_k in terms of the number of prime factors k

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p_k-1=\frac{\sigma(n)}{n}=\prod_{i=1}^{k}\left(1+\frac{1}{p_i}+\cdots+\frac{1}{p_i^{\alpha_i}}\right)<\prod_{i=1}^{k}\frac{p_i}{p_i-1}\le\prod_{i=1}^{k}\frac{i+1}{i}=k+1

Detailed analysis

Write $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ with $p_1<\cdots<p_k$ , so $p(n)=p_k$ . Using the multiplicative formula for $\sigma(n)$ , each factor $1+p_i^{-1}+\cdots+p_i^{-\alpha_i}$ is strictly less than the geometric series sum $p_i/(p_i-1)$ . Since $p_i\ge i+1$ and $t\mapsto t/(t-1)$ is decreasing, $\prod_{i=1}^k p_i/(p_i-1)\le\prod_{i=1}^k(i+1)/i=k+1$ , giving $p_k-1<k+1$ .