Are there $a,b \in \mathbb{N}$ that ${(\sum_{k=1}^n k)}^a = \sum_{k=1}^n k^b $ beside $2,3$

Solution 1:

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.

Solution 2:

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