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> There's also that a^3 + b^3 == c^3 problem that was solved in a few weeks by brute forcing every number (or something like that). Huh?
by jwilk 6y ago
> There's also that a^3 + b^3 == c^3 problem that was solved in a few weeks by brute forcing every number (or something like that).
Huh?
- analognoise 6y agoThere are only so many numbers, we simply tried them all. It took a really large computer. /s
- HPsquared 6y agoMaybe something to do with the taxicab numbers? https://en.wikipedia.org/wiki/Taxicab_number https://en.wikipedia.org/wiki/Taxicab_number https://mathworld.wolfram.com/TaxicabNumber.html https://mathworld.wolfram.com/TaxicabNumber.html
- Tainnor 6y agoYeah no idea, where they got that from. Fermat argued in the 1630s that this was impossible, Euler then proved it in the 18th century. [Actually, Fermat claimed that a^n+b^n=c^n is impossible for n>2, but the proof for that wasn't discovered until the 1990s. It's Andrew Wiles's famous proof of Fermat's Last Theorem.]
- dragontamer 6y agohttps://en.wikipedia.org/wiki/Sums_of_three_cubes https://en.wikipedia.org/wiki/Sums_of_three_cubes I was wrong and confused things for a different problem. a^3 + b^3 + c^3 == X, where X is some "hard" number, like 42. They discovered the a, b, and c values for X == 42 recently, by simplifying the problem over 2 variables and then brute forcing for weeks. ---------- Anyway, I was trying to reference the sum of 3-cubes issue.