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 3 of 3: Check k=1 and k=2 to isolate n=6

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k=1\Rightarrow\sigma(p^\alpha)=1+p+\cdots+p^\alpha\equiv1\pmod p\ \Longrightarrow\ n\nmid\sigma(n);\qquad k=2\Rightarrow n=2^\alpha3^\beta,\ \alpha=\beta=1\Rightarrow n=6

Detailed analysis

If $k=1$ then $n=p^\alpha$ and $\sigma(n)=1+p+\cdots+p^\alpha\equiv1\pmod p$ , so $n$ cannot divide $\sigma(n)$ , impossible. Hence $k=2$ , $p_1=2$ , $p_2=3$ , $n=2^\alpha3^\beta$ with $\alpha,\beta\ge1$ , and the equation requires $\sigma(n)/n=p_2-1=2$ . But $\sigma(n)/n=(1+\tfrac{1}{2}+\cdots+\tfrac{1}{2^\alpha})(1+\tfrac{1}{3}+\cdots+\tfrac{1}{3^\beta})\ge(1+\tfrac{1}{2})(1+\tfrac{1}{3})=\tfrac{3}{2}\cdot\tfrac{4}{3}=2$ , with equality if and only if $\alpha=\beta=1$ . Therefore $n=2^1\cdot3^1=6$ is the unique solution.