Are there $a,b \in \mathbb{N}$ that ${(\sum_{k=1}^n k)}^a = \sum_{k=1}^n k^b $ beside $2,3$ We know that:
$$\left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^n k^3 $$ 
My question is there other examples that satisfies: 
$$\left(\sum_{k=1}^n k\right)^a = \sum_{k=1}^n k^b $$
 A: Let $a=1+2+3+\cdots+n = \dfrac{n(n+1)} 2$.
Then whenever $p$ is an odd number, the sum
$$
1^p+2^p+3^p+\cdots + n^p
$$
can be written as a polynomial function of $a$.
https://en.wikipedia.org/wiki/Faulhaber%27s_formula#Faulhaber_polynomials
A: We can write
$$\sum^n_{k=1} k^a = \frac{1}{a+1} n^{a+1} + O(n^a).$$
The easiest way to convince yourself of this is to use a comparison test with an integral.
Thus, for the following to hold
$$\left(\sum_{k=1}^n k\right)^a = \sum_{k=1}^n k^b,$$
we need the highest order terms in the polynomials to be equal:
$$\left(\frac{1}{2}\right)^a n^{2a} = \frac{1}{b+1} n^{b+1}.$$
This gives the simultaneous set of equations
$$2^a=b+1$$
and
$$2a=b+1.$$
This can only hold when $a=2$, and hence $b=4-1=3$, or $a=1$ and $b=2-1=1$.
Certainly, we have $f(a)=2^a-2a$ for integer $a$ so $f''(a)=(\ln a)^2 2^a > 0$ and $f$ is strictly convex. It is easily shown that such a function cannot have more than two zeros immediately from the definition of convexity. Hence our two solutions are unique.
A: Put $n=2.$ Then the RHS is$$S=\sum k^b=2^b+1=\frac{2^{2b}-1}{2^b-1}.$$ By Zsigmondy's Theorem, if $b\ne 3$, there is a prime $p$ where the order of 2 mod $p$ is $2b$, and so $p\mid S$. (If $b=3$, the RHS is $9=3^2$ and there is no such $p$, for the order of 2 mod 3 is 2, not 6.)
Further, $p\mid (2^c+1)$ only if $b\mid c$, thus not for any $0<c<b$. But the only primes that divide $\left[\sum k\right]^a$ are those that divide $3^a$, viz 3.
Even if the LHS were generalised to $S=\left[\sum k^c\right]^a$, then, for a non-trivial solution, $a>1$ and $c<b$, so the above prime $p\nmid S$.
