No answers exist for a + b + c = 43 and a^3 + b^3 + c^3 = 17299^2,
even allowing them all to be negative:
http://ideone.com/FTm8l

On Sat, Jul 16, 2011 at 6:28 PM, Leopoldo Taravilse
<[email protected]> wrote:
> You can solve it with no programming.
> a+b+c = 43 && a^3 + b^3 + c^3 = 17299
> As 17299 == 3 mod 4 then as a^3,b^3 and c^3 can be 0 or 1 mod 4 then a, b
> and c are odd.
> 27^3 > 17299 so a < b < c <= 25
> 1^3 == 1 mod 5
> 2^3 == 3 mod 5
> 3^3 == 2 mod 5
> 4^3 == 4 mod 5
> 5^3 == 5 mod 5 (or 0^3 == 0 mod 5)
> So as 17299 == 43 + 1 mod 5 at least one of a, b and c must be 2 mod 5 so
> there are two options, 7 or 17.
> Case 1 (7): a+b = 36 && a^3 + b^3 = 16956
> Case 2 (17) a+b = 26 && a^3 + b^3 = 12386
> Let's see Case 1 first: a and b can be 1, 3, 5, 9, 11, 13, 15, 17, 19, 21,
> 23 or 25, but if a = 1, 3, 5 or 9 then b > 25 so a and b can be 11, 13, 15,
> 17, 19, 21, 23 or 25. We can easily discard 13 and 23 because of modulo 5
> reasons so we have 11, 15, 17, 19, 21 or 25 so we can try with this six
> numbers.
> Case 2: a and b can be 1, 3, 5, 9, 11, 13, 15, 17, 19, 21 or 23, but we can
> discard 3, 13 and 23 again so we have 1, 5, 9, 11, 15, 17, 19 and 21 to try.
>
> On Sat, Jul 16, 2011 at 6:54 AM, micke <[email protected]> wrote:
>>
>> What are the three positive integers whose sum is 43. and the sum of
>> the cubes of three integers is square of the number 17299.
>>
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