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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