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

Question

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 :)

Answer 1

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 $$ \frac{1}{2} \left( (y+1)^2 + (-x-1)^2 + (x-y)^2 \right)= \frac{1}{2} \left( (y+1)^2 + (x+1)^2 + (x-y)^2 \right) $$

Answer 2

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 $(a, b)=(-1,-1)$.