# A hard Cubic Diophantine equation

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.

• You might find something to note at matwbn.icm.edu.pl/ksiazki/aa/aa73/aa7331.pdf – Landuros Feb 19 '18 at 13:05
• Have a look here. – Dietrich Burde Feb 19 '18 at 13:09
• 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. – Tito Piezas III Feb 20 '18 at 10:37
• @user496634 Can you elaborate some? I don't get it :( – lminsl Oct 6 '18 at 7:10
• Sorry; I misread your question. – 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?
$x^3+y^3+z^3-2xyz=1$
Along with numerical solutions shown above $(x,y,z)=(1,1,1) = (13,13,-23)$ there is also the solution $(x,y,z)=( (468/659), (468/659), (-205/659 ) )$