A solution of Diophantine equation: $\big(x+y+z\big)^{3}=27x y z$ with $(x,y)∈Z$

Question

A solution is:

$x=(r+s)^{3}(r-3s)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$

$y=8r^{3}(2s-r)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$

$z=(r^{2}-2r s+3s^{2})^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$

Are there any more?

Answer 1

Let $x\to u+v,y\to u-v$, and let $u\to-\dfrac{T}{24z}-\dfrac{z}{2},v\to\dfrac{S}{216 z^2}$,

then $S^2 = (T + 9 z^2) (T + 36 z^2)^2 \implies T + 9 z^2 = \square$

gp-code:

xyz()=
{
 for(u=1, 1000,
  T= thue('x^2+3, 48*u^2);
  for(i=1, #T,
   q= T[i][1];
   z= T[i][2]-4*u;
   if(z,
    t= -12*z*(2*u+z);
    s= q*(t+36*z^2);
    v= s/216/z^2;
    x= u+v; y= u-v;
    if(x & y,
     if(x==floor(x) & y==floor(y),
      if(gcd(x,y)==1 & gcd(x,z)==1 & gcd(y,z)==1,
       if((x + y + z)^3 == 27*x*y*z,
        kx= ispower(x)%3==0 || abs(x)==1;
        ky= ispower(y)%3==0 || abs(y)==1;
        kz= ispower(z)%3==0 || abs(z)==1;
        print("("x", "y", "z")    ",kx&ky&kz)
       )
      )      
     )
    )
   )
  )
 )
};

Some solutions:

(1, 1, -8)    1
(-1, 27, -8)    1
(27, -1, -8)    1
(1, 27, -64)    1
(27, 1, -64)    1
(-27, 125, -8)    1
(125, -27, -8)    1
(-1, 125, -64)    1
(125, -1, -64)    1
(1, 125, -216)    1
(125, 1, -216)    1
(27, 125, -512)    1
(125, 27, -512)    1
(-125, 343, -8)    1
(343, -125, -8)    1
(-27, 343, -64)    1
(343, -27, -64)    1
(-1, 343, -216)    1
(343, -1, -216)    1
(1, 343, -512)    1
(343, 1, -512)    1
(27, 343, -1000)    1
(343, 27, -1000)    1
(-343, 729, -8)    1
(729, -343, -8)    1
(125, 343, -1728)    1
(343, 125, -1728)    1
(-729, 1331, -8)    1
(1331, -729, -8)    1
(-125, 729, -64)    1
(729, -125, -64)    1
(-1, 729, -512)    1
(729, -1, -512)    1
(1, 729, -1000)    1
(729, 1, -1000)    1
(125, 729, -2744)    1
(729, 125, -2744)    1
(-1331, 2197, -8)    1
(2197, -1331, -8)    1
(-343, 1331, -64)    1
(1331, -343, -64)    1
(343, 729, -4096)    1
(729, 343, -4096)    1
(-2197, 3375, -8)    1
(3375, -2197, -8)    1
(-125, 1331, -216)    1
(1331, -125, -216)    1
(-27, 1331, -512)    1
(1331, -27, -512)    1
(-1, 1331, -1000)    1
(1331, -1, -1000)    1
(1, 1331, -1728)    1
(1331, 1, -1728)    1
(27, 1331, -2744)    1
(1331, 27, -2744)    1
(125, 1331, -4096)    1
(1331, 125, -4096)    1
(-729, 2197, -64)    1
(2197, -729, -64)    1
(-3375, 4913, -8)    1
(4913, -3375, -8)    1
(343, 1331, -5832)    1
(1331, 343, -5832)    1
(-343, 2197, -216)    1
(2197, -343, -216)    1
(-4913, 6859, -8)    1
(6859, -4913, -8)    1

Yes, all $x,y,z$ is cubes.

Answer 2

Once we have proved that all of $x,y,z$ are, if the gcd is 1, cubes themselves, we arrive at, using $$ \delta = \frac{-1 + i \sqrt 3}{2} $$ so that $\delta^2 + \delta + 1 = 0$ and $\delta^3=1,$

$$ ( u^3 + v^3 + w^3)^3 - 27 u^3 v^3 w^3 = \left( \color{red}{u^3 + v^3 + w^3 - 3uvw} \right) \left( \color{blue}{u^3 + v^3 + w^3 - 3 \delta uvw} \right) \left( \color{green}{u^3 + v^3 + w^3 - 3 \delta^2 uvw} \right) $$

Let me post a clean proof that, if $\gcd(x,y,z) = 1$ and $(x+y+z)^3 = 27 xyz,$ then actually $\gcd(x,y) = \gcd(y,z) = \gcd(z,x) = 1,$ therefore $x,y,z$ are all cubes.

We demand, by dividing through if necessary, that $\gcd(x,y,z) = 1.$ Let us ASSUME that there are a pair that are not coprime, let those be $y,z.$ Next, let $g = \gcd(y,z)$ and ASSUME $g > 1.$ Next we get $y = g \beta$ and $z = g \gamma,$ with $\gcd(\alpha, \beta) = 1.$ The important part is that, from the hypothesis $\gcd(x,y,z) = 1,$ actually $$ \gcd(x,g) = 1. $$ Pick some new letters, $$ \color{magenta}{mx + ng = 1.} $$

The diophantine equation becomes $$ 27x g^2 \beta \gamma = (x + g \beta + g \gamma)^3 = x^3 + g \cdot (\mbox{stuff}) $$ Thus we have $$ g | x^3 \; . \; \; $$ Well, $$g | (m x^3 + ng x^2) = x^2 (mx+ng) = x^2$$ so $g | x^2.$ Well, $$g | (m x^2 + ng x) = x (mx+ng) = x$$ so $g | x.$ Well, $$g | (m x + ng ) = 1$$ so $g | 1.$ Then $$ g = 1 $$ which contradicts the assumption that there is a pair of coprime variables among $x,y,z.$ The contradiction proves that $\gcd(x,y) = \gcd(y,z) = \gcd(z,x) = 1.$ As we have $xyz = t^3$ for some integer $t,$ we find that $x,y,z$ are all cubes. Taking $x = u^3, y = v^3, z=w^3,$ the identity

$$ ( u^3 + v^3 + w^3)^3 - 27 u^3 v^3 w^3 = \left( \color{red}{u^3 + v^3 + w^3 - 3uvw} \right) \left( \color{blue}{u^3 + v^3 + w^3 - 3 \delta uvw} \right) \left( \color{green}{u^3 + v^3 + w^3 - 3 \delta^2 uvw} \right) $$

says that either $x=y=z$ or $x+y+z=0.$

Answer 3

$\big(x+y+z\big)^{3}=27xyz\tag{1}$
Let $s=x+y+z$ and substitute $z=s-x-y$ to equation $(1).$
$s^3-27xys+27x^2y+27xy^2=0\tag{2}$
For $x$ to be rational solution, the discriminant of the equation $(2)$ must be square.
Hence $-27y(4s-3y)(s-3y)^2$ must be square, then
$v^2 = (9y-6s)^2-36s^2$
We solve simultaneous equations $[6s=m^2-n^2,9y-6s=m^2+n^2].$
We get $y = \frac{2m^2}{9}, s = \frac{m^2-n^2}{6}, x = \frac{-(m-n)^3}{36m}, z=\frac{-(m+n)^3}{36m}.$
Finally, $(x,y,z)=(-(m-n)^3, 8m^3, -(m+n)^3).$ $(m,n)=(2,1): [x,y,z]=[-1, 64, -27]$
$(m,n)=(3,2): [x,y,z]=[-1, 216, -125]$
$(m,n)=(4,1): [x,y,z]=[-27, 512, -125]$
$(m,n)=(4,3): [x,y,z]=[-1, 512, -343]$

Added new solutions:

We can get new solution using a known solution of equation $(1).$
$(x,y,z)=(-(m-n)^3, 8m^3, -(m+n)^3)$ is a known solution.
Substitute $x = t-(m-n)^3, y = t+8m^3, z = kt-(m+n)^3$ to equation $(1)$, and let $k= \frac{-(5m^4+8m^3n+2m^2n^2+n^4)}{4(m^2-2mn+n^2)m^2},$ then $t = \frac{-72(m^4-2m^3n+2n^3m-n^4)m^3}{9m^4-18m^3n-6n^3m-n^4}.$
Hence we get new parametric solution below.
$(x,y,z)=(-(3m+n)^3(m-n)^3, -64n^3m^3, (m+n)^3(-n+3m)^3).$

Similarly, we get other new solution using above new solution below. $(x,y,z)=(-(m-n)^3(3m+n)^3(-n^2+6mn+3m^2)^3, 512(3m^2-n^2)^3n^3m^3, (-n+3m)^3(m+n)^3(-n^2-6mn+3m^2)^3).$

Answer 4

We have the identity:

$a^3+b^3+c^3=3abc$

Condition is, $(a+b+c)=0$

In "OP" solution if we take,

$a=(r+s)(r-3s)$

$b=2r(2s-r)$

$c=(r^2-2rs+3s^2)$

We get, $(a+b+c)=0$

Hence, $a^3+b^3+c^3=3abc$ ----(1)

Therefore by cubing both sides of equation (1),

$(a^3+b^3+c^3)^3=(3abc)^3=27(abc)^3 =27(a^3)(b^3)(c^3)$

Let, $(x,y,z)=(a^3,b^3,c^3)$

And we get:

$(x+y+z)^3=27xyz$

The constant, $w=(r^4-4r^3s+4r^2s^2+3s^4)$

is a multiplication factor & can be multiplied in if required.