Sum of three cubes for 42 finally solved

[Albert Chan]

(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
b = { 80435758145,  817515}   -- b^3 ~ +5.2041E+50
c = { 12602123297,  335631}   -- c^3 ~ +0.0200E+50

function cube(x, n)
    x = (x[1]%n * 1e6 + x[2]) % n
    return x * x % n * x % n
end

modtest = function(n) return (cube(a,n) + cube(b,n) + cube(c,n)) % n end 


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