Diophantine equation with cubes.

Question

Interested in the solution in general Diophantine equations of the form:

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

$q$ - what some integer.

Solutions similar equations can be written.

Since this equation is easy, as it is quite symmetrical.

$X^3+Y^3+Z^3-3XYZ=R^3$

Such a solution can write.

$X=sp(p+s)$

$Y=s(2p^2+s^2)$

$Z=p(p^2+2s^2)$

$R=p^3+s^3$

And solutions can be written:

$X=s(9p^2+9ps+10s^2)$

$Y=s(6p^2+12ps+7s^2)$

$Z=3p^3+3p^2s+15ps^2+7s^3$

$R=3(p+2s)(p^2-ps+s^2)$

Whether there are any thoughts how to solve this equation?

At first I thought to use for solving Pell's equation, but I think that you can do without.

Answer 1

You can use the following identity:

$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$$

Fixed integer q has a finite number of factorial expansions. So we get a finite number of systems of two equations:

$$x+y+z = a$$ $$x^2+y^2+z^2-xy-xz-yz = b$$

for all integers $a$ and $b$ such that $ab=q$. Excluding the variable z we obtain the quadratic equation in x and y.

CALCULATIONS:

Let $u = x + y$ and $v=x y$. Then we have $z = a - u$ and $$x^2+y^2+z^2-xy-xz-yz = u^2 - 2v + (a-u)^2 - v - (a-u)u =$$ $$u^2 - 3v + a^2 - 2au + u^2 - au +u^2 = 3u^2 - 3au - 3v + a^2$$

Thus, we have a equation in integers u and v: $$ 3(u^2 - au) - 3v = b - a^2$$ It`s something like Pell's equation as you said. But it is more simple and we can express v from u:

$$ 3v = 3(u^2 - au) - (b - a^2)$$

Next step. Let suppose that (x,y) is an integer solution of original equation then because $u = x + y$ and $v=x y$ there is integer m such that $u^2-4v=m^2$ (this is a discriminant of a square equation $(\lambda - x)(\lambda - y) = \lambda^2-u\lambda+v=0$).

So $4v = u^2 - m^2$ and we have the following transformations of the equation: $$ 3v = 3(u^2 - au) - (b - a^2)$$ $$12(u^2 - au)-3*4v = 4(b-a^2)$$ $$12(u^2 - au)-3(u^2-m^2)=4(b-a^2)$$ $$9u^2 - 12au + 3m^2=4(b-a^2)$$ $$9u^2 - 12au + 4a^2 + 3m^2=4b$$ $$(3u-2a)^2 + 3m^2=4b$$

Thus, we have that the number of solution for each fixed a and b is finite! So for each parameter q we can find all solutions (x,y,z) of the original equation $x^3+y^3+z^3 = 3xyz + q$ by the following formulas:

$$x=\frac{u-m}{2}$$ $$y=\frac{u+m}{2}$$ $$z=a-u$$

where u and m form a solution of the equation:

$$(3u-2a)^2 + 3m^2=4b$$

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.