Team Selection Test

The answer is $a=3,b=1,c=2$ and $a=b\in {\mathbb{Q}}^{+},c=1$. Applying induction on $n$ by using Bernoulli's inequality gives $$\frac{n^{r+1}}{r+1}\le S_r(n)\le \frac{(n+1{)}^{r+1}}{r+1}$$ for all positive integer $n$ and positive rational number $r$. As $S_a(n)=(S_b(n){)}^c$ letting $r=a$ and $r=b$ gives that $$\frac{n^{a+1}}{a+1}\le {\left(\frac{(n+1{)}^{b+1}}{b+1}\right)}^c\text{ and }{\left(\frac{n^{b+1}}{b+1}\right)}^c\le \frac{(n+1{)}^{a+1}}{a+1}$$ i.e. $$\frac{n^{(b+1)c}}{(n+1{)}^{a+1}}\le \frac{(b+1{)}^c}{a+1}\le \frac{(n+1{)}^{(b+1)c}}{n^{a+1}}$$ holds for infinitely many positive integers $n$. By letting $n\to \infty$ in the last inequality, we obtain that $(b+1)c=a+1$ and $(b+1{)}^c=a+1$. If $c=1$, then $a=b$ and we get the trivial solutions. If $c>1$, then $c=(b+1{)}^{c-1}$ implies that $b$ is an integer since $c$ is an integer. As $b\ge 1$, we get that $c\ge 2^{c-1}$ and hence $c=2$. This leads to $b=1,a=3$ and this solution clearly satisfies the condition.