# x^3 + y^3 = 35

• Oct 5th 2006, 06:05 AM
beta12
x^3 + y^3 = 35
Find all the solutions in positive integers to : x^3 + y^3 = 35 .

Can you teach me how to solve this question? Thank you very much.
• Oct 5th 2006, 06:14 AM
topsquark
Quote:

Originally Posted by beta12
Find all the solutions in positive integers to : x^3 + y^3 = 35 .

Can you teach me how to solve this question? Thank you very much.

Well, the numbers are small enough for trial and error. The only other way I can think of would be to factor:

x^3 + y^3 = (x + y)(x^2 + xy + y^2) = 35.

The only integer factors of 35 are (1, 35) or (5, 7). There are no integers x and y that add to 1, so x + y = 5 or 7.

Using x + y = 5 we get that y = 5 - x, so y^3 = 125 - 75x + 15x^2 - x^3 so
x^3 + y^3 = 15x^2 - 75x + 125 = 35
or
15x^2 - 75x + 90 = 0
x^2 - 5x + 6 = 0
(x - 3)(x - 2) = 0
So x = 2 or x = 3
Thus y = 3 or y = 2 respectively.

Using x + y = 7 we get that y = 7 - x, so y^3 = 343 - 147x + 21x^2 - x^3 so
x^3 + y^3 = 21x^2 - 147x + 343 = 35
21x^2 - 147x + 308 = 0
3x^2 - 21x + 44 = 0
or
x = (21 +/- sqrt(-87))/6
Since x is complex, this can't be a solution.

So (x, y) = (2, 3) or (x, y) = (3, 2)

However, this is only going to be practical when there are a small number of possibilities for x + y.

-Dan
• Oct 5th 2006, 07:44 AM
beta12
Hi topsquark,

Thank you very much. I solved this question easily with your guideline.

This is a typo in your factorization. It should be
x^3 + y^3 = (x + y)(x^2 - xy + y^2) = 35.
• Oct 5th 2006, 07:45 AM
topsquark
Quote:

Originally Posted by beta12
Hi topsquark,

Thank you very much. I solved this question easily with your guideline.

This is a typo in your factorization. It should be
x^3 + y^3 = (x + y)(x^2 - xy + y^2) = 35.

:o Oops! Thanks for the catch!

-Dan