Cube which can be Represented as Sum of 3, 4, 5, 6, 7 or 8 Cubes
Theorem
Proof
\(\ds 351120^3\) | \(=\) | \(\ds 175560^3 + 234080^3 + 292600^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \times 87780^3 + 204820^3 + 321860^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \times 87780^3 + 175560^3 + 2 \times 263340^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3 \times 117040^3 + 3 \times 234080^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \times 58520^3 + 117040^3 + 3 \times 175560^3 + 292600^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 8 \times 175560^3\) |
These representations are not necessarily unique.
$\blacksquare$
Additional Results
We also have:
\(\ds 351120^3\) | \(=\) | \(\ds 58520^3 + 2 \times 117040^3 + 5 \times 175560^3 + 234080^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 5 \times 87780^3 + 4 \times 175560^3 + 263340^3\) |
So $351120^3$ can also be expressed as a sum of $9$ or $10$ cubes.
These equations all stem from:
\(\ds 12^3\) | \(=\) | \(\ds 6^3 + 8^3 + 10^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \times 3^3 + 7^3 + 11^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \times 3^3 + 6^3 + 2 \times 9^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3 \times 4^3 + 3 \times 8^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \times 2^3 + 4^3 + 3 \times 6^3 + 10^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 8 \times 6^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^3 + 2 \times 4^3 + 5 \times 6^3 + 8^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 5 \times 3^3 + 4 \times 6^3 + 9^3\) |
showing that $351 \, 120$ is not the smallest number with this property.
Moreover, using $3^3 + 4^3 + 5^3 = 6^3$, this result could still be further extended:
\(\ds 12^3\) | \(=\) | \(\ds 2^3 + 3^3 + 3 \times 4^3 + 5^3 + 4 \times 6^3 + 8^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 6 \times 3^3 + 4^3 + 5^3 + 3 \times 6^3 + 9^3\) | ||||||||||||
\(\ds \) | \(:\) | \(\ds \) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 9 \times 3^3 + 4 \times 4^3 + 4 \times 5^3 + 9^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^3 + 5 \times 3^3 + 7 \times 4^3 + 5 \times 5^3 + 8^3\) |
which is a sum of $11$ to $19$ cubes.
Using $1^3 + 6^3 + 8^3 = 9^3$ and $\paren {2 n}^3 = 8 \times n^3$, we can express $12^3$ as more and more cubes.
Historical Note
This result can be found in Curious and Interesting Numbers by David Wells from $1986$, but it is a mystery as to why it was presented.
There appears to be absolutely nothing special about $351 \, 120$, as there are many cubes far smaller which possess the same property.
It is possible that Wells misinterpreted something that caught his eye, but googling for properties of $351 \, 120$ reveals that it is spectacularly uninteresting.
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $351,120$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $351,120$