Cubic diophantine equation with a prime $x^3 + y^3 + z^3 - 3xyz = p$.

Question

Question: Find all triple positive integers $(x, y, z)$ so that $$x^3 + y^3 + z^3 - 3xyz = p,$$ where $p$ is a prime number greater than $3$.

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I have tried the following: The equation factors as

$$(x + y + z) (x^2 + y^2 + z^2-xy-yz-zx) = p.$$

Since $x + y + z> 1$, we must have $x + y + z = p$ and $$x^2 + y^2 + z^2-xy-yz - zx = 1.$$ The last equation is equivalent to $$(x-y)^2 + (y-z)^2 + (z-x)^2 = 2.$$ Without loss of generality you can assume that $x ≥ y ≥ z$, we have $xy ≥ 1$ and $xz ≥ 2$, implying $$(xy)^2 + (yz)^2 + (zx)^2 ≥ 6> 2.$$

Who can help me and correct me, thank you.

Answer 1

Result: If $p>3$ is a prime number and $x$, $y$ and $z$ are positive integers such that $$x^3 + y^3 + z^3 - 3xyz = p,$$ then if $p\equiv1\pmod{3}$ we have, after permuting $x$, $y$ and $z$, that $$(x,y,z)=\left(\tfrac{p-1}{3},\tfrac{p-1}{3},\tfrac{p+2}{3}\right),$$ and if $p\equiv2\pmod{3}$ we have, after permuting $x$, $y$ and $z$, that $$(x,y,z)=\left(\tfrac{p+1}{3},\tfrac{p+1}{3},\tfrac{p-2}{3}\right).$$

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Proof: As you already note, the equation can be expressed as $$(x + y + z) (x^2 + y^2 + z^2-xy-yz-zx) = p,$$ which immediately shows that, because $x$, $y$ and $z$ must be positive, $$x+y+z=p\qquad\text{ and }\qquad x^2 + y^2 + z^2-xy-yz-zx=1.\tag{1}$$ The latter can be rewritten as $$(x-y)^2+(x-z)^2+(y-z)^2=2,$$ which shows that two of the three numbers are the same, and the third differs from them by only $1$. That is to say, without loss of generality we have $$x=y=z\pm1.$$ Plugging this back into the first equation found at $(1)$ shows that $$p=x+y+z=3x\pm1,$$ and so we find that $x=\tfrac{p\mp1}{3}$. As $x$ must be an integer we see that only one of the two choices of the $\pm$-sign is possible, depending on whether $p\equiv1\pmod{3}$ or $p\equiv2\pmod{3}$.

Conversely, a routine check shows that if $p\equiv\pm1\pmod{3}$ then the triplet of positive integers $$(x,y,z)=\left(\tfrac{p\mp1}{3},\tfrac{p\mp1}{3},\tfrac{p\pm2}{3}\right),$$ and its three distinct permutations satisfy the equation $$x^3 + y^3 + z^3 - 3xyz = p.$$

Answer 2

Strangely enough, the solution is finite.

for the equation:

$$X^3+Y^3+Z^3-3XYZ=q=ab$$

If it is possible to decompose the coefficient as follows: $4b=k^2+3t^2$

Then the solutions are of the form:

$$X=\frac{1}{6}(2a-3t\pm{k})$$

$$Y=\frac{1}{6}(2a+3t\pm{k})$$

$$Z=\frac{1}{3}(a\mp{k})$$

Thought the solution is determined by the equation Pell, but when calculating the sign was a mistake. There's no difference, but the amount should be. Therefore, the number of solutions of course.

I may be wrong, though. We still need to check other options.