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

Answer 3

I have another solution but it is not short as the previous ones.

First of all, we see that $(-1, -1)$ is a solution. Next, we can consider the cubic curve $f(x, y)=x^3+y^3+3xy-1=0$ for $(x, y)\in \mathbb{R}^2$ and show that $(-1, -1)$ is a singular point of multiplicity 2. It holds, $f_x=3x^2+3y=0$ and $f_y=3y^2+3x=0$ if and only if $(x, y)=(-1, -1)$ or $(x, y)=(0, 0)$ but $(0, 0)$ does not belong the cubic curve. Now, we know that $(-1, -1)$ is a double point. Hence, the line with slop $m$ through $(-1, -1)$ has intersection multiplicity of 2 with the curve. I. e., the polynomial $p(x)=f(x, m(x+1)-1)$ of degree 3 has a double root at $x=-1$. We can do polynomial division to find the third root of $p$. We get after some long calculations that $x=\dfrac{2-m}{m+1}$ and $y=m(x+1)-1=m\left(\dfrac{2-m}{m+1}+1\right)-1=\dfrac{2m-1}{m+1}$ for $m\neq -1$. Adding $x$ and $y$ yields $$x+y=\dfrac{2-m}{m+1}+\dfrac{2m-1}{m+1}=1.$$ Now, the solution set over integers is given by $$ \left\{(x, y)\in \mathbb{Z}^2: x+y=1\right\}\cup \left\{(-1, -1)\right\}. $$ Note, that for $m=-1$ the line is parallel to the curve restricted for $(x, y)\in \left\{(x, y)\in \mathbb{R}^2: x+y=1\right\}$. (Remark: $(-1, -1)$ is an isolated point of the cubic curve with multiplicity 2 and therefore $f(x, y)$ must have a linear factor; in our case it is $y+x-1$.)