# Integer solutions for a simple cubic

What are the integer solutions to $$x^3 - 7xy + y + 1 = 0$$? A computation only finds $$(0, -1), (-1, 0), (-6, 5), (-49, 342)$$. This is surprisingly few.

Are these all of them? Is there an algebraic geometry proof? Finite fields? Is this much harder than I realize?

• The equation is equivalent to $\frac{x^3+1}{7x-1}=y$. Commented Jul 25 at 5:33

There are no other solutions.

As Lucenaposition observed, we are seeking integers $$x$$ such that $$y(x) := (x^3+1) / (7x-1)$$ is an integer. Expanding about $$x=\infty$$ we find that $$343 \, y(x) = 49 x^2 + 7 x + 1 + \frac{344}{7x-1}$$ so if both $$x$$ and $$y$$ are integers then so is $$344 / (7x-1)$$. That is, $$7x-1$$ must be a factor of $$344 = 2^3 43$$. The only factors congruent to $$-1 \bmod 7$$ are $$-1, -8, -43, -344$$. These correspond exactly to the solutions $$x=0, -1, -6, -49$$ that WSJ already found.

• What does expanding about $x=\infty$ mean? Commented Jul 25 at 21:30
• Doesn't this imply then that integer solutions to similar equations are no more than the number of divisors of the (integral) numerator of the remainder? ie that there are few is not surprising at all? Commented Jul 26 at 13:46