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Prove that the only integer solutions to $x^3 + 3x^2y - 3xy^2 - 3y^3 = 1$ are $(1,0)$ and $(-2,3)$.

My only idea for now is to try to represent this in the form $A^3 + B^3 = C^3$ and apply Fermat's Last Theorem. However, I cannot find suitable $A,B,C$.

It is also easy to see that firstly $x\equiv 1 \pmod 3$ and then $3$ divides $y$, but substituting does not give a better looking equation.

Another alternative: if we set $s = x+y$, then we get $s^3 -6sy^2 + 2y^3 = 1$ which looks a bit better.

Any help appreciated!

DesmondMiles
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  • Dividing everything by $y^3$ looks nice but I haven't been able to get anywhere. – Toby Mak Mar 14 '20 at 06:26
  • This Thue equation. In pari/gp thue(x^3+3*x^2-3*x-3,1), then output [[-2, 3], [1, 0]]. – Dmitry Ezhov Mar 14 '20 at 06:58
  • Ok but that is cheating. Any idea how such things are proven? – DesmondMiles Mar 14 '20 at 07:04
  • @DesmondMiles, x=-2, y=3 does not satisfy the equation, so the statement of question is wrong. the only solution is x=1 and y=0. – sirous Mar 14 '20 at 09:46
  • @sirous x=-2, y=3 does satisfy the equation – miracle173 Mar 15 '20 at 03:07
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    @DesmondMiles xould you provide more context? – miracle173 Mar 15 '20 at 03:11
  • Maybe could be useful to write the given equation in this equivalent form: $,(s-1)\cdot(s^2+s+1)=2y^2(3s-y)$, with $,s=x+y$. Clearly, if $,s,$ is even the equation has no solution. So, we can define $,k=\frac{s-1}2,$ and write the equation as $,k(3+6k+4k^2)=y^2(3+6k-y),$. – Augusto Santi Mar 15 '20 at 23:46
  • @DesmondMiles, the equation can be written as $(x+3y)(x^2-3y^2)=1-6y^3$,If y=3 then RHS can be factorized as $3^4--3^4)(3^4-1+3^4)=-1(2\times 3^4-1)=-161=(7)(-23)$, which gives $x=2$. If $x=1$ it can easily be seen that y=0. – sirous Mar 16 '20 at 12:57

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