This problem is from an Olympiad handout:

Show that there exists infinitely many integer triplets $(x,y,z)$ such that $ x^3+y^3+z^3-2xyz=1$.

I tried to plug $x=y$ and use tangent lines to find solutions inductively, but the method didn't work well. (It gave $(1,1,1) \rightarrow (13,13,-23)$ ,but after that there was only rational roots)

Also I plugged the equation in the cubic formula and tried to delete the cubic root, but it also failed.

Help me please :I

  • 1
    $\begingroup$ You might find something to note at matwbn.icm.edu.pl/ksiazki/aa/aa73/aa7331.pdf $\endgroup$ – Landuros Feb 19 '18 at 13:05
  • $\begingroup$ Have a look here. $\endgroup$ – Dietrich Burde Feb 19 '18 at 13:09
  • $\begingroup$ There are infinitely many rational solutions using an elliptic curve. However, the cubic Pell has form, $$x^3+dy^3+d^2z^3-3dxyz = 1$$ which is not the form of your equation, so it may be doubtful it has infinitely many integer solutions. And a quick computer search doesn't find even find "smallish" others except those already given. $\endgroup$ – Tito Piezas III Feb 20 '18 at 10:37
  • $\begingroup$ @user496634 Can you elaborate some? I don't get it :( $\endgroup$ – lminsl Oct 6 '18 at 7:10
  • $\begingroup$ Sorry; I misread your question. $\endgroup$ – YiFan Oct 6 '18 at 8:43

This isn't a full solution, but it's a start. Take modulo $x$, so $y^3+z^3\equiv1$. Set $y^3+z^3=ax+1$. Similarly, set $x^3+z^3=by+1$ and $x^3+y^3=cz+1$. Then $$x^3+y^3+z^3=\tfrac12(ax+by+cz+3).$$ Now express $2xyz$ in terms of $(a,b,c)$ by solving the three simultaneous equations. Namely, $x=\sqrt[3]{\frac12(-ax+by+cz+1)}$ And similarly for the other variables, so $2xyz=\sqrt[3]{(-ax+by+cz+1)(ax-by+cz+1)(ax+by-cz+1)},$ Noting that the factor of $2$ cancels out. Now the expression inside the root must be a perfect cube. Maybe you can try continuing from here?


Above equation shown below,


Along with numerical solutions shown above $(x,y,z)=(1,1,1) = (13,13,-23)$ there is also the solution $ (x,y,z)=( (468/659), (468/659), (-205/659 ) )$


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