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.
 A: 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 $n=1$.
A: Your statement is false. If you use $n=1$, both sides of your statement equal $1$. For that matter, if $n=0$ then both sides are zero, but you may not consider zero to be a natural number.
It is true that the statement is true for $n>1$. Is that what you want to prove? If so, find the polynomial expressions for each side. Subtracting one side from the other leads you to search for the roots of a sixth-degree polynomial. Show that two of the roots are $0$ and $1$, three roots are negative, and one root is between zero and one. That accounts for all six roots, so there is no root for $n>1$. Or, as @Deepak suggests, use the rational root theorem to find all rational roots, which are just $-1,0,1$.
A: As Rory Daulton has remarked we have $L(0)=R(0)$ and $L(1)=R(1)$. But then $L(2)=17< 27=R(3)$. Therefore it is sufficient to prove
$$L(n)-L(n-1)\leq R(n)-R(n-1)\qquad(n\geq3)\ ,$$
which is the same as
$$n^4\leq \left({n(n+1)\over2}\right)^3-\left({(n-1)n\over2}\right)^3=n^3\ {6n^2+2\over 8}\qquad(n\geq3)\ .$$
This amounts to $4n\leq3n^2+1$, resp., to  $$(3n-1)(n-1)\geq0\qquad(n\geq3)\ ,$$
which  is obviously true.
