crazyfehmy wrote: Let $S_r(n)=1^r+2^r+\cdots+n^r$ where $n$ is a positive integer and $r$ is a rational number. If $S_a(n)=(S_b(n))^c$ for all positive integers $n$ where $a, b$ are positive rationals and $c$ is positive integer then we call $(a,b,c)$ as nice triple. Find all nice triples. Asymptotically, $S_r(n)$ grows like $n^{r+1}/(r+1)$. Hence $S_a(n)=(S_b(n))^c$ yields $\displaystyle \lim_{n\to\infty} \left(n^{a+1}/(a+1)\right) /\left(n^{b+1}/(b+1)\right)^c=1$, which implies $a+1=(b+1)c$ and $a+1=(b+1)^c$. Hence $(b+1)c=(b+1)^c$ and $c=(b+1)^{c-1}$ with a positive rational $b$ and a positive integer $c$. For $c=1$ we get $a=b$; this yields the nice triples $(k,k,1)$ with $k$ a positive integer. For $c\ne1$, we get from $c=(b+1)^{c-1}$ that $b+1$ must be integer. Since $b+1\ge2$, we further get that $c\ge2^{c-1}$ and hence $c=2$. This leads to $b=1$ and $a=3$, and yields the nice triple $(3,1,2)$.

test20 wrote: crazyfehmy wrote: Let $S_r(n)=1^r+2^r+\cdots+n^r$ where $n$ is a positive integer and $r$ is a rational number. If $S_a(n)=(S_b(n))^c$ for all positive integers $n$ where $a, b$ are positive rationals and $c$ is positive integer then we call $(a,b,c)$ as nice triple. Find all nice triples. Asymptotically, $S_r(n)$ grows like $n^{r+1}/(r+1)$. Hence $S_a(n)=(S_b(n))^c$ yields $\displaystyle \lim_{n\to\infty} \left(n^{a+1}/(a+1)\right) /\left(n^{b+1}/(b+1)\right)^c=1$, which implies $a+1=(b+1)c$ why ???

Asymptotically, $S_r(n)$ grows like $n^{r+1}/(r+1)$. By taking the integral, but is an approximation. Hence $S_a(n)=(S_b(n))^c$ yields $\displaystyle \lim_{n\to\infty} \left(n^{a+1}/(a+1)\right) /\left(n^{b+1}/(b+1)\right)^c=1$, which implies $a+1=(b+1)c$ and $a+1=(b+1)^c$. Taking $n$ to $n+1$ and getting eq. it would mean $(\frac{n+1}{n})^i$ is at both sides equal, so the $i$ is the same, but cause $n \to infty$ and it was an approximation, I'm not sure if we may say this all. IMPORTANT: taking $b \sim 0.495, c=5,a\sim 6.17$ we get $90 $ vs $80$ for $n=2.$ Hence your way isn't correct, you have luck in natural numbers, it is correct here, but is a wrong proof.

SCP wrote: Taking $n$ to $n+1$ and getting eq. it would mean $(\frac{n+1}{n})^i$ is at both sides equal, so the $i$ is the same, but cause $n \to infty$ and it was an approximation, I'm not sure if we may say this all. IMPORTANT: taking $b \sim 0.495, c=5,a\sim 6.17$ we get $90 $ vs $80$ for $n=2.$ Hence your way isn't correct, you have luck in natural numbers, it is correct here, but is a wrong proof. I do not have the slightest idea what you want to say. Perhaps you are mixing up the if-part and the then-part of my if-then statement? I am using the following statement: Conditions: Let $f_1(n)$, $f_2(n)$, $g_1(n)$, $g_2(n)$ be functions from the positive integers into the positive integers, that are increasing and tend to infinity, such that $f_1$ grows asymptotically like $g_1$ and $f_2$ grows asymptotically like $g_2$. Statement: If $f_1(n)=f_2(n)$ for all $n$, then $\lim_{n\to\infty} g_1(n)/g_2(n)=1$.

Given that order of S_r(n) is O(n^(r+1)), we get a+ 1 = (b+1).c Remark that S_r(n) / n^(r+1) tends to integral from 0 to 1 of f(x) = x^(r+1)/(r+1) Using equation a+1 = (b+1).c with the above remark we get 1/(a+1) = (1/(b+1))^ c

Here is a complete solution by using Bernoulli's inequality. The only possible triples are $(a,b,c)=(3,1,2)$ and $(a,b,c)=(k,k,1)$ where $k$ is a positive raitonal. Claim: For every positive integer $n$ and positive raitonal $r$, \[ n^{r+1}/(r+1) \leq S_r(n) \leq (n+1)^{r+1}/(r+1) \]Proof: Just apply induction on n by using Bernoulli's inequality. Hence $S_a(n)=(S_b(n))^c$, letting $r=a$ and $r=b$ gives that, \[ n^{a+1}/(a+1) \leq ((n+1)^{b+1}/(b+1))^c\]\[ {and} \]\[(n+1)^{a+1}/(a+1) \geq (n^{b+1}/(b+1))^c \]If we multiply these inequalities we get \[ n^{(b+1)\cdot (c)}/(n+1)^{a+1} \leq (b+1)^c/(a+1) \leq (n+1)^{(b+1)\cdot (c)}/n^{a+1} \]holds for every positive integer $n$. By letting $n\to\infty$ we get that, $a+1=(b+1)c$ and $a+1=(b+1)^c.$ If $c=1$, then $a=b.$ If $c\geq 2$, then $c=(b+1)^{c-1}$ implies $b$ is an integer. So $b\geq 1 \implies c\geq 2^{c-1} \implies c=2$ which gives us $(a,b,c)=(3,1,2).$