An exercise from Math StackExchange

Problem:

Give all the positive whole number solutions to the equation $x^3−y^3 = 602$

Solution:

Here is really just a replication of my answer on math.stackexchange.com:

Note that $(x - y)^2 = x^2 - 2xy + y^2$, we can write

$x^3 - y^3 = (x - y)(x^2 + xy + y^2) = (x - y)((x - y)^2 + 3xy)$

For simplicity, let $z = x - y$, we have

$602 = z(z^2 + 3xy)$

Suppose for a moment that $z$ is known, now we can calculate

$z^2 + 3xy = \frac{602}{z}$

$3xy = \frac{602}{z} - z^2$

$3(x - y + y)y = \frac{602}{z} - z^2$

$3(z + y)y = \frac{602}{z} - z^2$

$3zy + 3y^2 = \frac{602}{z} - z^2$

$3y^2 + 3zy + z^2 - \frac{602}{z} = 0$

Despite the deceiving complexity, since $z$ is assumed to be known, we can easily find $y$ using the quadratic formula.

Now we have $602 = 2 \times 7 \times 43$, so $z$ can only be these options

* 1
* 2
* 7
* 43
* $2 \times 7$
* $2 \times 43$
* $7 \times 43$
* $2 \times 7 \times 43$

And the negative of these values

Out of these 16 choices, we can easily enumerate the solutions. Of course, many of these choices does not generate integer solution, just ignore them.

For example, if I choose $z = 2$, we get $11^3 - 9^3 = 602$ and also $(-9)^3 - (-11)^3 = 602$.