How to prove a sum of series How do I prove that for any natural number $n$ we have
$$\sum_{i=0}^n i^4 \neq \left(\sum_{i=0}^n i\right)^3?$$
Any help would be greatly appreciated.

Recall that $$\sum_{i=1}^n i^4 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
and
$$\sum_{i=1}^n i = \dfrac{n(n+1)}2$$
We hence need
$$\left(\dfrac{n(n+1)}2\right)^3 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
This gives us either $n(n+1)=0$ or
$$15(n(n+1))^2 = 4(2n+1)(3n^2+3n-1) \,\,\,\, (\spadesuit)$$
$(\spadesuit)$ can be simplified as
$$15n^4+6n^3-21n^2-4n+4 = 0 \implies (n-1)(15n^3+21n^2-4) = 0 \,\,\,\, (\clubsuit)$$
Now consider the function $f(n) = (15n^3+21n^2-4)$. Note that we have $f'(n) = 45n^2+42n >0 $ for all $n \geq 1$. Hence, the function $f(n)$ is increasing for all positive integers. Further, $f(1) = 32 > 0$. Hence, $f(n)$ has no non-negative integer as its roots. The only solution to your initial problem is $n=0$ and $n=1$.
