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I've tried the following:

$x^3-y^3 = (x-y)(x^2+xy+y^2) = 602.$ That's it, I'm stuck.

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    Hint - there is only a handful of possible factorization of 602 – 2017-01-11
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    Also both $x,y$ will have to be odd and you can also say $\gcd(x,y)=1$ – 2017-01-11
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    You have a typo in the factorization. The first term should be $x-y$ – 2017-01-11

3 Answers 3

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Hint: $602 \equiv 2 \pmod 8$. Hence $(x-y) \equiv 2 \pmod 8$.

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meanwhile, if they are integers and not equal, $$ |x-y| \geq 1. $$ In any case, $$ x^2 + xy + y^2 \geq \frac{3}{4} \; x^2, $$ $$ x^2 + xy + y^2 \geq \frac{3}{4} \; y^2. $$

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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 $.