Solve $a^3+b^3+3ab=1$ with $(a,b)\in \Bbb{Z}^2$

Solve the following equation for $$(a,b)\in \Bbb{Z}^2$$: $$a^3+b^3+3ab=1$$

I tried all of the standard techniques I know. I tried modular arithmetic:

$$a^3+b^3+3ab\equiv 1 \pmod{3}$$ $$a^3+b^3\equiv 1 \pmod{3}$$

Now by Fermat's Little Theorem:

$$a^2 a+b^2 b\equiv 1 \pmod{3}$$ $$a+b\equiv 1 \pmod{3}$$

But I can't see the next move I have to do. I can't find any banal factorization of the first term(it would require solving a $$3$$ degree equation). I tried using classic scomposition such as the sum of $$2$$ cubes and the cube of a binomial. Thank you for your time :)

• See here. Apr 18, 2019 at 19:02
• $(a,b)=(2,-1)$ is a solution, so trying to prove it has no solutions will fail. Apr 18, 2019 at 19:06
• @DietrichBurde and how should I come up with such a scomposition Apr 18, 2019 at 19:07

The item worth memorizing is $$x^3 + y^3 + z^3 - 3xyz = (x+y+z)\left( x^2 + y^2 + z^2 - yz - zx - xy \right)$$ where the quadratic form is positive semidefinite because $$\left( x^2 + y^2 + z^2 - yz - zx - xy \right) = \frac{1}{2} \left( (y-z)^2 + (z-x)^2 + (x-y)^2 \right)$$
If you then take $$z=-1$$ you get $$x^3 + y^3 -1 + 3xy = (x+y-1)\left( x^2 + y^2 +1 + y + x - xy \right)$$ and the quadratic factor is $$\frac{1}{2} \left( (y+1)^2 + (-x-1)^2 + (x-y)^2 \right)= \frac{1}{2} \left( (y+1)^2 + (x+1)^2 + (x-y)^2 \right)$$
The equation is equivalent to $$(a+b-1)\left(\left(a-\frac{b-1}{2}\right)^2+\frac{3}{4}(b+1)^2\right)=0.$$ So either $$a+b-1=0$$, or $$(a, b)=(-1,-1)$$.
• The idea is to factorize. Once we have the factor $a+b-1$, it is easy. The factorisation can be seen by observing that $b=1-a$ works. Apr 18, 2019 at 19:12