Sum of three cubes for 42 finally solved
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09-15-2019, 04:20 PM
(This post was last modified: 09-15-2019 04:39 PM by Albert Chan.)
Post: #7
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RE: Sum of three cubes for 42 finally solved
(09-14-2019 09:14 PM)Gerson W. Barbosa Wrote: 34 significant digits are not enough for the task. That would require at least 50. We can split the numbers, like this: a = −80538738812075974 = −80538739e9 + 187924026 = a1 + a0 b = +80435758145817515 = +80435758e9 + 145817515 = b1 + b0 c = +12602123297335631 = +12602123e9 + 297335631 = c1 + c0 a³ + b³ + c³ = (a1 + a0)³ + (b1 + b0)³ + (c1 + c0)³ = (a1³ + b1³ + c1³) + 3*(a0a1² + b0b1² + c0c1²) + 3*(a1a0² + b1b0² + c1c0²) + (a0³ + b0³ + c0³) = -6628842934503040000000000000000000000000000 +6628842934562563667007015000000000000000000 -59523703031106040674633000000000 +36024091040674633000000042 = 42 For calculators with precision of 12 decimal digits, we can do mod test: PHP Code: a = {-80538738812, -075974} -- a^3 ~ -5.2241E+50 lua> modn = {999907,999917,999931,999953,999959,999961,999979,999983,1e6} lua> for _, n in ipairs(modn) do print(modtest(n), '(mod ' .. n .. ')') end 42 (mod 999907) 42 (mod 999917) 42 (mod 999931) 42 (mod 999953) 42 (mod 999959) 42 (mod 999961) 42 (mod 999979) 42 (mod 999983) 42 (mod 1e+006) Except for 1e6, all mods are primes, we get a³ + b³ + c³ ≡ 42 (mod ~ 1e54) Since 1e54 ≫ |sum of three cubes| , we get a³ + b³ + c³ = 42 |
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