On the equation $x^3+y^3+z^3-2xyz=N$

The following question was asked at MSE without any solution:

Show that the equation $x^3+y^3+z^3-2xyz=1$ have infinitely many integer solutions $(x,y,z)$.

A more general question was also proposed at MSE:

For which positive integers $N$ does the equation $x^3+y^3+z^3-2xyz=N$ have infinitely many integers solutions $(x,y,x)$?

Some references were given, like the solutions of $x^3+y^3+z^3=nxyz$.

Usually cubic diophantine equations are very hard to deal with.

Does anyone know how any of these two problems could be solved?
Is the second problem (on the equation $x^3+y^3+z^3-2xyz=N$) open?

Any help would be appreciated.

• I am guessing the original intention was to state a cubic norm form equation, where the norm form is the principal norm form of a cubic field (so that it represents one). Then the result would be obvious from the fact that the unit group of a ring of integers in a cubic field is always infinite. The second question seems very hard. For instance, it is not known exactly which integers are sums of three cubes. – Stanley Yao Xiao Mar 24 '18 at 14:59
• @StanleyYaoXiao: The OP and you claim that there are infinitely many integer solutions of $x^3+y^3+z^3−2xyz=1$. It does not seem to be true. Are there integer solutions except $(1,1,1), (1,0,0),(13,13,−23)$ (and permutations of those)? The question itself is a bad copy of the question from MSE linked there. I voted to close. – Mark Sapir Mar 25 '18 at 13:53
• @Mark Sapir I don’t claim that this equation has infinitely many solutions, but say $x^3 + 2y^3 + 4z^3 - 6xyz = 1$ definitely has infinitely many solutions, since the left hand side is the principal norm form of the cubic field $\mathbb{Q}(\sqrt{2})$, and there is a non-trivial unit group which acts on the solutions, generating infinitely many. This was the intent of my first comment – Stanley Yao Xiao Mar 25 '18 at 16:06
• @StanleyYaoXiao: OK, I understand your comment now. It is easy to see that if 1 is replaced by, say, 19, then there are infinitely many integer solutions of the original equation. – Mark Sapir Mar 25 '18 at 16:11