Can an integer be decomposed as $\sum_{k=1}^{m}k^3$? (and how to find $m$?)

Question

I have two questions:

1. I wonder how to check if an integer $n$ can be represented as:

$$n = \sum_{k=1}^{m}k^3$$

2. And if it can be decomposed this way, how to find $m$?

I tried to use an integral approximation like this: $k = \sqrt[3]{4x}$, but it is not correct, because I need an accurate (descrete) answer.

Answer 1

Since $$\sum\limits_{k=0}^m k^3=\left(\frac{m(m+1)}{2}\right) ^2$$ (you can easily prove this by induction), $n$ necessarily has to be a perfect square $i^2$ because $\frac{m(m+1)}{2}\in\mathbb{N}$. Then we have $$m(m+1)=2i\Leftrightarrow m^2+m-2i=0.$$ If this equation has a positive integer root $m_1$, then $$\sum\limits_{k=0}^{m_1} k^3=n.$$ If such a root does not exist, the number $n$ cannot be represented in the desired way.

Answer 2

Just to show how to prove the formula $\sum_{k=1}^n k^3=\frac14n^2(n+1)^2$ with no previous knowledge.

Let $f:\Bbb R\to \Bbb R$ be a function such that $f(x)=f(x-1)+x^3$ and $f(0)=0$. Assume for the moment that $f$ is a polynomial of degree $4$ like $f(x)=ax^4+bx^3+cx^2+dx$ (note that since $f(0)=0$, $f$ has no constant term).

Then $$ax^4+bx^3+cx^2+dx=a(x-1)^4+b(x-1)^3+c(x-1)^2+d(x-1)+x^3$$ So $$b=-4a+b+1$$ $$c=6a-3b+c$$ $$d=-4a+3b-2c+d$$ It is a triangular system. Solve it and you are almost done.

Now that you have the formula, you can prove it by induction.