$20x^3 - 14y^2 = 2014$

The solution has to be a pair of integers (x, y). I've been trying to solve this one for the past few hours. I've brute forced pairs of values where $x$ and $y$ are in the range of $[1, 10000]$ or where $x$ is in the range of $[1, 1000000]$ with no success.

Can you help me solve it? Is there a general solution for equations that have the form $ax^3-by^2=10a + b$?


  • $\begingroup$ Do your answers need to be integers, real numbers, complex? $\endgroup$ – Rocket Man Jun 2 '14 at 15:54
  • $\begingroup$ Sorry, forgot to mention that they have to be integers :) $\endgroup$ – nimsquier Jun 2 '14 at 15:55

Write $a=x^{3},b=y^{2}$ then you get $20a+14b=2014$ which is $10a+7b=1007$. Then using well known method to generalize from particular solution to general solution which result in $a=100+7k,b=1-10k$. Now using modulo we get $x^{3}\equiv 2(\mod 7)$ which is clearly absurb. Hence no solution exist.

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  • $\begingroup$ You could've checked the equation directly mod 7. +1 though for the result. $\endgroup$ – Macavity Jun 2 '14 at 16:13
  • $\begingroup$ @Gina, The most tricky part is : choosing the modulus.How have you identified $7$ as the mod? $\endgroup$ – lab bhattacharjee Jun 2 '14 at 16:17
  • $\begingroup$ A good motivation for $7$ would be Fermat's Little Theorem leaving only $3$ possibilities for $x^3$, and the fact that $14y^2$ cancels. $\endgroup$ – Bart Michels Jun 2 '14 at 16:24

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