Diophantine problem: Find all pairs of $(x, y)$ such that $x^3-4xy+y^3=-1$

contest-math elementary-number-theory diophantine-equations

Solution 1:

Hint: A messy quadratic is better than a simple cubic while dealing with integers. $$x+y=s\implies x^3-4x(s-x)+(s-x)^3=-1 $$

Simplifying, $$ (3s+4)x^2-(3s^2+4s)x+s^3+1=0$$

The discriminant must be non-negative, $$\Delta=-(3s+4)(s^3-4s^2+4)\ge 0$$

If $s\in (-\infty,-2]$ or $s\in [4,\infty)$, $\Delta<0$. Therefore, $s\in\{-1,2,3\}$. Only $s=-1$ results in an integer value for $x$.

$$(x,y)\in \{(-1,0),(0,-1)\}$$

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