Property of 490,689
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Theorem
The number $490 \, 689$ can be expressed as the sum of $3$ cubes in $2$ different ways:
- $490 \, 689 = 4^3 + 60^3 + 65^3 = 8^3 + 25^3 \times 78^3$
while at the same time the products of the contributory cube roots of each of those $2$ ways are equal:
- $4 \times 60 \times 65 = 8 \times 25 \times 78$
Proof
\(\ds 490 \, 689\) | \(=\) | \(\ds 64 + 216 \, 000 + 274 \, 625\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4^3 + 60^3 + 65^3\) |
\(\ds 490 \, 689\) | \(=\) | \(\ds 512 + 15 \, 625 + 474 \, 552\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 8^3 + 25^3 + 78^3\) |
Then:
\(\ds 15 \, 600\) | \(=\) | \(\ds 2^4 \times 3 \times 5^2 \times 13\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^2 \times \left({2^2 \times 3 \times 5}\right) \times \left({5 \times 13}\right)\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4 \times 60 \times 65\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^3 \times 5^2 \times \left({2 \times 3 \times 13}\right)\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 8 \times 25 \times 78\) |
$\blacksquare$
Historical Note
This result is reported by David Wells in his Curious and Interesting Numbers, 2nd ed. of $1997$ as the work of Stephane Vandemergel, but details are lacking.
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $490,689$