# All integer solutions of $x^3-y^3=2020$.

Find all integer pairs $$(x,y)$$ satisfying $$x^3-y^3=2020\,.$$

First, $$x^3-y^3=(x-y)(x^2+xy+y^2)=2020$$ and $$2020=2^2\cdot 5 \cdot 101$$.

But what next? Can it be worked out by using modulo? Or how? Any idea? Thanks in advance.

There are no solutions, because $$x^3,y^3\equiv0$$ or $$\pm1\pmod7$$, but $$2020\equiv4\bmod7$$.

• I could have said $9$ instead of $7$ – J. W. Tanner Jun 3 at 14:31
• Both are motivated by $\phi(9) = \phi(7) = 6$, which is a small multiple of $3$, so there will be few values of $x^3$ mod $7$ or mod $9$. – Misha Lavrov Jun 3 at 14:33
• What mathematical principle did you utilise here? I’ve never learned about this kind of math – gen-z ready to perish Jun 3 at 14:53

Let $$d = x - y$$. Then we want $$x^3 - (x-d)^3 = 2020$$, which is a quadratic equation in $$x$$. The discriminant is $$24240d - 3d^4 = 3d (8080 - d^3)$$, which is nonnegative only for $$d \in [0, 8080^{1/3}]$$, and since $$d$$ is an integer it must be between $$0$$ and $$20$$.

Moreover, since $$(x-y)^3 = d(x^2 + xy + y^2) = 2020$$, $$d$$ must be a divisor of $$2020$$. This leaves only $$6$$ possibilities: $$d = 1, 2, 4, 5, 10, 20$$. We already had a finite problem to solve before, but this observation reduces the number of cases.

For each value of $$d$$, the solutions to the quadratic equation have irrational solutions, so there are no integer solutions $$(x,y)$$.

Alternatively, if $$p$$ is a prime natural number that divides $$x^2+xy+y^2$$, then $$(2x+y)^2+3y^2=4(x^2+xy+y^2)\equiv 0\pmod{p}\,.$$ Thus, either $$p$$ divides both $$x$$ and $$y$$, or $$\left(\dfrac{-3}{p}\right)=1$$. Now, by quadratic reciprocity, $$1=\left(\dfrac{-3}{p}\right)=\left(\dfrac{p}{-3}\right)=\left(\dfrac{p}{3}\right)\,,$$ whence $$p\equiv 1\pmod{3}$$. Because $$(x-y)(x^2+xy+y^2)=x^3-y^3=2020=2^2\cdot 5\cdot 101$$ with $$5\not\equiv 1\pmod{3}$$ and $$101\not\equiv 1\pmod{3}$$, we conclude that $$5$$ and $$101$$ cannot divide $$x^2+xy+y^2$$. Thus, the only possible prime divisor of $$x^2+xy+y^2$$ is $$2$$, and if $$2\not\equiv 1\pmod{3}$$ is a factor of $$x^2+xy+y^2$$, we must have $$2\mid x$$ and $$2\mid y$$. Since $$x^2+xy+y^2\geq 0$$, this implies $$x^2+xy+y^2=1\text{ or }x^2+xy+y^2=4\,.$$ The only solutions $$(x,y)\in\mathbb{Z}\times\mathbb{Z}$$ to $$x^2+xy+y^2=1$$ are $$(x,y)=\pm (1,0),\pm(0,1),\pm(1,-1)\,.$$ The only solutions $$(x,y)\in\mathbb{Z}\times\mathbb{Z}$$ to $$x^2+xy+y^2=4$$ are $$(x,y)=\pm (2,0),\pm(0,2),\pm(2,-2)\,.$$ None of these solutions satisfies $$x^3-y^3=2020$$.