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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 :)

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

2 Answers 2

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

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

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  • $\begingroup$ What's the idea behind this scomposition? $\endgroup$
    – Kandinskij
    Apr 18, 2019 at 19:11
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    $\begingroup$ 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. $\endgroup$ Apr 18, 2019 at 19:12

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