Korean Mathematical Olympiad

We use an induction on $n$.

a. For $n=6$, put $N_6=2\cdot 3\cdot 7\cdot 43\cdot 1807=1806\cdot 1807$. Since $1807=13\cdot 139$, $N_6$ is divisible by 6 distinct primes. Moreover, since $$\frac{1}{m}=\frac{1}{m+1}+\frac{1}{m(m+1)},$$ we have

$$\begin{array}{rl}1& =\frac{1}{2}+\frac{1}{2}\\ & =\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\\ & =\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}\\ & =\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}+\frac{1}{1806},(42\cdot 43=1806)\\ & =\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}+\frac{1}{1807}+\frac{1}{N_6}.\end{array}$$ Hence, by multiplying both sides of above equality by $N_6$, we get $$N_6=1+\frac{N_6}{2}+\frac{N_6}{3}+\frac{N_6}{7}+\frac{N_6}{43}+\frac{N_6}{1807},$$ where each term of the right hand side is a divisor of $N_6$.

b. Suppose there is an $n$-good number $N_n$. Put $$N_{n+1}=N_n(N_n+1).$$ Then $$\begin{array}{rl}{N}_{n+1}& =(1+x_2+\cdots +x_n)(N_n+1)\\ & =1+N_n+x_2(N_n+1)+\cdots +x_n(N_n+1).\end{array}$$ Hence $N_{n+1}$ is a sum of $n+1$ distinct divisors. Since $(N_n,N_n+1)=1$, prime divisors of $N_n+1$ are different from those of $N_n$. Since $N_n+1$ has at least one prime divisor, and since $N_n$ has at least $n$ distinct prime divisors, $N_{n+1}$ has at least $n+1$ distinct prime divisors. Therefore $N_{n+1}$ is a $(n+1)$-good number. $\square$