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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.

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

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    $\begingroup$ I could have said $9$ instead of $7$ $\endgroup$ – J. W. Tanner Jun 3 at 14:31
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    $\begingroup$ 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 $9$. $\endgroup$ – Misha Lavrov Jun 3 at 14:33
  • $\begingroup$ What mathematical principle did you utilise here? I’ve never learned about this kind of math $\endgroup$ – gen-z ready to perish Jun 3 at 14:53
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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 $(x,y)$.

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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 $x^3-y^3=2020$.

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