P.S.:

http://www.wolframalpha.com/input/?i=real+solutions+%28a^3+%2B+b^3+%2B+%2843+-+a+-+b%29^3+%3D+17299^2%29<http://www.wolframalpha.com/input/?i=real+solutions+%28a%5E3+%2B+b%5E3+%2B+%2843+-+a+-+b%29%5E3+%3D+17299%5E2%29>

<http://www.wolframalpha.com/input/?i=real+solutions+%28a%5E3+%2B+b%5E3+%2B+%2843+-+a+-+b%29%5E3+%3D+17299%5E2%29>
http://www.wolframalpha.com/input/?i=integer+solutions+%28a^3+%2B+b^3+%2B+%2843+-+a+-+b%29^3+%3D+17299^2%29<http://www.wolframalpha.com/input/?i=integer+solutions+%28a%5E3+%2B+b%5E3+%2B+%2843+-+a+-+b%29%5E3+%3D+17299%5E2%29>

http://www.wolframalpha.com/input/?i=rational+solutions+%28a^3+%2B+b^3+%2B+%2843+-+a+-+b%29^3+%3D+17299^2%29<http://www.wolframalpha.com/input/?i=rational+solutions+%28a%5E3+%2B+b%5E3+%2B+%2843+-+a+-+b%29%5E3+%3D+17299%5E2%29>


On Sat, Jul 16, 2011 at 15:54, Diego Gavinowich <[email protected]> wrote:

> There are infinite (real) solutions for:
>
> a^3 + b^3 + (43 - a - b)^3 = 17299^2
>
> Basically you can replace *b* with any number and you have a quadratic
> equation (because a^3 will go away)...
> Any quadratic equation has 2 solutions (might give the same value, and both
> are complex or real).
>
> -3ba^2 + 129a^2 - 3ab^2 + 258ab - 5547a + 129b^2 - 5547b - 299175894 = 0
>
> If *b* was a constant then we have:
>
> a^2(-3b+129) + a (-3b^2+258b-5547) + (129b^2 - 5547b - 299175894) = 0
>
> (-3b^2+258b-5547)^2 - 4 * (-3b+129) * (129b^2 - 5547b - 299175894) >= 0 is
> what we need to have both real solutions (if I didn't make any silly
> mistake) and that has infinite solutions (*b* <= 43 or *b* >= ~722.943).
>
> Best,
> Diego
>
>
>
> On Sat, Jul 16, 2011 at 15:32, Alfonso J. Ramos <[email protected]> wrote:
>
>> Excuse me (I used spanish): not in the integers, but apparently there is
>> one in the reals:
>>
>> http://www.wolframalpha.com/input/?i=+a+%2B+b+%2B+c+%3D+43+and+a^3+%2B+b^3+%2B+c^3+%3D+17299^2<http://www.wolframalpha.com/input/?i=+a+%2B+b+%2B+c+%3D+43+and+a%5E3+%2B+b%5E3+%2B+c%5E3+%3D+17299%5E2>
>>
>> 2011/7/16 Alfonso J. Ramos <[email protected]>
>>
>> No en los enteros, pero al parecer si en los reales:
>>>
>>> http://www.wolframalpha.com/input/?i=+a+%2B+b+%2B+c+%3D+43+and+a^3+%2B+b^3+%2B+c^3+%3D+17299^2<http://www.wolframalpha.com/input/?i=+a+%2B+b+%2B+c+%3D+43+and+a%5E3+%2B+b%5E3+%2B+c%5E3+%3D+17299%5E2>
>>>
>>>
>>> 2011/7/16 Luke Pebody <[email protected]>
>>>
>>>> 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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>>>>
>>>
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