Cubed triangle

$136 \times 46 = 6256 = 10^3 + 11^3 + 12^3 + 13^3$ is the sum of $\boxed{\text{four}}$ consecutive cubes.

This problem was inspired by the fact that the sum of the first $n$ consecutive cubes equals the square of the $n$ th triangle number. $C_n = 1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2 = T_n^2,$ If $m < n$ we also have $C_n - C_m = T_n^2 - T_m^2 = (T_n + T_m)(T_n - T_m).$ This shows that any sum of consecutive cubes can be written as the product of the sum and the difference of two triangle numbers.

For instance, $28$ and $10$ are the 7th and 4th triangle numbers. Thus we have $(28+10)(28-10) = 28^2 - 10^2 = (1^3 + \cdots + 7^3) - (1^3 + \cdots + 4^3) = 7^3 + 6^3 + 5^3.$ Likewise, $91$ and $45$ are the 13th and 9th triangle numbers, so that $(91+45)(91-45) = 91^2 - 45^2 = (1^3 + \cdots + 13^3) - (1^3 + \cdots + 9^3) = 13^3 + 12^3 + 11^3 + 10^3.$ The quick way to answer the question is therefore to observe that the difference between 13 and 9 is equal to 4.

Conversely, we now have a non-trivial factoring for any sum of consecutive cubes. E.g. $8^3 + 9^3 + 10^3 + 11^3 + 12^3 + 13^3 = 7497 = 119 \times 63 = (91 + 28)(91 - 28).$

The $n$ th triangular number is given by $\displaystyle T_n = \sum_{k=1}^n k = \frac {n(n+1)}2$ . Consider the following product:

$\begin{aligned} P(m,n) & = (T_n + T_m)(T_n - T_m) & \small \color{#3D99F6} \text{where }n > m. \\ & = T_n^2 - T_m^2 \\ & = \left(\frac {n(n+1)}2\right)^2 - \left(\frac {m(m+1)}2\right)^2 & \small \color{#3D99F6} \text{Note that }\sum_{k=1}^n k^3 = \left(\frac {n(n+1)}2\right)^2 \\ & = \sum_{k=1}^n k^3 - \sum_{k=1}^m k^3 \\ & = \sum_{k=m+1}^n k^3 & \small \color{#3D99F6} \text{The sum of }n-m \text{ consecutive cubes.} \end{aligned}$

This means that $P(m,n)$ is the sum of $n-m$ consecutive cubes. SInce we can find $n$ of $T_n$ from $n = \left \lfloor \sqrt {2T_n} \right \rfloor$ , where $\lfloor \cdot \rfloor$ denotes the floor function , then $(91+ 45)(91-45)$ is equal to the sum of $\left \lfloor \sqrt {2\times 91} \right \rfloor - \left \lfloor \sqrt {2\times 45} \right \rfloor = 13-9 = \boxed 4$ consecutive cubes.

I have done it using my scientific calculator

It can be observed that the sum of two consecutive triangular numbers T(n-1) and T(n) is n-squared while the difference is n. The product is therefore n-cubed.

Now express the product of the sum and difference of any two numbers is equal to the difference of their squares. (A+B)(A-B) = A^2-B^2. It's a relatively simple step now to express the product of the sum and difference of any two triangular numbers as the total of the product-sum-difference of the each consecutive triangular numbers in between. These are the consecutive cubes we require.

!0^3 + 11^3 +12^3 + 13^3 = 6256. Ed Gray