you are right luke.

well thanks for trying.

On Jul 16, 10:50 am, Luke Pebody <[email protected]> wrote:
> 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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