Sum of odd powers

I would like to offer a non-inductive proof which hopefully explains how Jerry Han Jia Tao arrived at his initial equation.

Firstly, by trying small cases, we note that f(n) is in fact 0. We are then motivated to prove that the formula for sum of fifth powers contains the same factors as the formula for cubes. The formula for cubes is more well-known and can be done in the same way as for the sum of fifth powers, so its derivation will not be shown in this solution.

From the formulas for sum of n natural numbers and sum of n squares, we see that the degree is 1 more than the power the terms are raised to.

Hence, let $\displaystyle \sum_{n=0}^x n^5 = h(x) = ax^6+bx^5+cx^4+dx^3+ex^2+fx+g$

Since $h(x)$ is the sum of fifth powers to $x$ , we expect $h(x+1)-h(x)=(x+1)^5$ . Also, we extend h(x) to $x=0$ so that we can see $g = 0$ .

Now, $h(x+1) = a(x+1)^6+b(x+1)^5+c(x+1)^4+d(x+1)^3+e(x+1)^2+f(x+1)+g$ $=ax^6 + (6a+b)x^5+(15a+5b+c)x^4+(20a+10b+4c+d)x^3+(15a+10b+6c+3d+e)x^2+(6a+5b+4c+3d+2e+f)x+(a+b+c+d+e+f+g)$

The coefficients are easily obtained from Pascal's Triangle , so expansion is not too difficult.

Subtracting h(x), $h(x+1)-h(x) =6ax^5+(15a+5b)x^4+(20a+10b+4c)x^3+(15a+10b+6c+3d)x^2+(6a+5b+4c+3d+2e)x+(a+b+c+d+e+f)$

Also, $(x+1)^5 = x^5+5x^4+10x^3+10x^2+5x+1$

Comparing coefficients, we obtain a system of equations:

$6a=1 \Rightarrow a=\frac {1}{6}$

$15a+5b=5 \Rightarrow b=\frac {5-15a}{5} \Rightarrow b=\frac {1}{2}$

$20a+10b+4c=10 \Rightarrow c=\frac {10-20a-10b}{4} \Rightarrow c=\frac {5}{12}$

$15a+10b+6c+3d=10 \Rightarrow d=\frac {10-15a-10b-6c}{3} \Rightarrow d=0$

$6a+5b+4c+3d+2e=5 \Rightarrow e=\frac {5-6a-5b-4c-3d-2e}{2} \Rightarrow e=-\frac {1}{12}$

$a+b+c+d+e+f=1 \Rightarrow f=1-a-b-c-d-e \Rightarrow f=0$

$\Rightarrow (a,b,c,d,e,f,g)=(\frac {1}{6},\frac {1}{2},\frac {5}{12},0,-\frac {1}{12},0,0)$

Now to factorise the equation into $\frac {1}{6}x^6+\frac {1}{2}x^5+\frac {5}{12}x^4-\frac {1}{12}x^2$ $=\frac {1}{12}x^2(2x^4+6x^3+5x^2-1)$ $=\frac {1}{12}x^2(x+1)(2x^3+4x^2+x-1)$ $=\frac {1}{12}x^2(x+1)^2(2x^2+2x-1)$

as desired. We factored out the $(x+1)^2$ because the formula for sum of cubes $(\frac {x(x+1)}{2})^2$ also contains the $(x+1)^2$ which we hope to cancel out.

It can now be seen that $\frac {3(\frac {1}{12}x^2(x+1)^2(2x^2+2x-1))}{(\frac {x(x+1)}{2})^2}$ $=2x^2+2x-1$ which is an integer for integer values of x.

The required answer is then $0+0+\dots + 0 = \boxed{0}$