How many integer solutions are there to $x^3-y^3=271$.

  • 8
    $\begingroup$ Hint: $271$ is prime and $x^3-y^3 = (x-y)(x^2+xy+y^2)$. $\endgroup$
    – Arthur
    Mar 4, 2015 at 17:13

1 Answer 1


I claim only two such that $x,y \in \mathbb Z$.

Indeed, it is clear that $x>y$, we say that $x+i=y$ where $i>0$, $i \in \mathbb Z$, then $(y+i)^3-y^3=3iy^2+3i^2y+i^3=271$, moreover $3y^2+3iy+i^2=\frac{271}{i}$ is integer because $y$ is supposed to be integer and $i$ is integer, $271$ is prime, thus $i$ must be $1$ or $271$.

If $i=1$, we need to find an integer solution to $3y+3y^2+1=271$, there are only two $y=-10$ and $y=9$.

If $i=271$, then $3y^2+3iy+i^2=\frac{271}{i}$ does not have real roots.

Thus $y=-10$ $x=-9$ and $y=9$ and $x=10$ are the only solutions.

  • $\begingroup$ You're missing the case $i=271$. $\endgroup$
    – kingW3
    Mar 4, 2015 at 17:37
  • $\begingroup$ @kingW3 fixed it in the solution, thanks. $\endgroup$
    – zesy
    Mar 4, 2015 at 17:41

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