When you don't start at 1

For positive integer $n$ , define $T_n = 1+2+\cdots + n$ and $C_n = 1^3+2^3+\cdots + n^3$ .

The identity we're given at the start is $C_n=T_n^2$ , which holds for all $n$ .

To neaten up the notation, put $c=a-1$ . We're told $c>0$ .

The equation can then be written as

$(T_b-T_c)^2=C_b-C_c=T_b^2-T_c^2$

Since $b>a>c$ we can divide through by $T_b-T_c$ to give

$T_b-T_c=T_b+T_c$

which can only happen when $T_c=0$ ; but this is impossible, since $c>0$ ; so there are no other pairs of positive integers that satisfy the equation.

The $\text{LHS}$ of the last "equation" can be expressed as:

$\begin{aligned} S & = \left(\sum_{k=a}^b k \right)^2 \\ & = \left(\sum_{k=1}^b k - \sum_{k=1}^{a-1} k \right)^2 \\ & = \left(\sum_{k=1}^b k \right)^2 - 2\sum_{k=1}^b k \sum_{k=1}^{a-1} k + \left(\sum_{k=1}^{a-1} k \right)^2 \\ & = \sum_{k=1}^b k^3 - 2\sum_{k=1}^b k \sum_{k=1}^{a-1} k + \sum_{k=1}^{a-1} k^3 \end{aligned}$

If $\text{LHS} = \text{RHS}$ , then

$\begin{aligned} \sum_{k=1}^b k^3 - 2\sum_{k=1}^b k \sum_{k=1}^{a-1} k + \sum_{k=1}^{a-1} k^3 & = \sum_{k=a}^b k^3 \\ & = \sum_{k=1}^b k^3 - \sum_{k=1}^{a-1} k^3 \\ \implies \sum_{k=1}^b k \sum_{k=1}^{a-1} k & = \sum_{k=1}^{a-1} k^3 \end{aligned}$

The above equation holds true only when $b=1$ and $a=2$ , which contradicts with the assumption of $b>a$ . Therefore, there are no other pairs of $(a,b)$ that the equation holds.