How to show that if $x, y, z$ are rational numbers satisfying $(x + y + z)^3 = 9(x^2y + y^2z +z^2x)$, then $x = y = z$

Question

Let $x,y,z$ rationals
Show that if
$(x+y+z)^3=9(x^2y+y^2z+z^2x)$ then $x=y=z$

I tried this : Let $x$ be the smallest variable
Write $y=a+x$ and $z=b+x$
Prove $a=b=0$ by factoring the equation as the sum of three squares. any suggestions?

Answer 1

Since this is a homogeneous polynomial equation with cyclic symmetry, WLOG divide out by $z^3$ and set $x'\leftarrow x/z$, $y'\leftarrow y/z$. Then we're left with the affine cubic curve $$(x'+y'+1)^3=9(x'^2 y' + y'^2+x')\text{.}$$ Then rational solutions for $(x',y')$ correspond to rational homogeneous solutions $[x:y:z]$ of the original equation. Certainly $(x',y')=(1,1)$ satisfies this equation, so our strategy is to expand around this point: set $x\leftarrow 1+u$, $y\leftarrow 1+v$. Expanding and simplifying (disclosure: I used a CAS) gives the homogeneous cubic equation $$u^3-6u^2v+3uv^2+v^3=0$$ (geometrically, the tangent cone of the affine curve at $(1,1)$). This is a homogeneous equation: rational solutions for $(x',y')$ above correspond with rational homogeneous solutions for $[u:v]$ here. We see that $v=0$ implies $u=0$, so assume that $v\neq 0$, divide by $v^3$, and set $u'\leftarrow u/v$. Then we're tasked with solving the affine cubic equation $$u'^3-6u'^2+3u'+1=0\text{.}$$ Rational solutions for $u'$ here correspond with rational homogeneous solutions for $[u:v]$ above. By the Rational Root Theorem, any rational value for $u'$ must satisfy $u'=\pm 1$. But neither of these is a root, so there are no nontrivial rational solutions $u'$. Reversing the correspondences, there are no nontrivial rational solutions for $[u:v]$, $(x',y')$, and finally $[x:y:z]$.

("Nontrivial" $[x:y:z]$ means in addition $[x:y:z]\neq[1:1:1]$.)

Answer 2

CONCLUSION:The roots of $\eta^3 - 3 \eta - 1 = 0$ are $$ A = 2 \cos \left( \frac{7 \pi}{9} \right) \approx -1.532 \; \; \; , B = 2 \cos \left( \frac{5 \pi}{9} \right) \approx -0.347 \; \; \; , C = 2 \cos \left( \frac{ \pi}{9} \right) \approx 1.879 \; \; \; . $$ Note that we do have $$ A+B+C=0 \; . \; $$ We get identity $$ \color{magenta}{ (Ax+By+Cz)(Bx+Cy+Az)(Cx+Ay+Bz) = (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right)} $$ which confirms that the surface is three planes sharing the line $x=y=z\; . \;$

ORIGINAL:

Take $$ x = r-s-t \; , \; \; \; y = r+s-t \; , \; \; \; z = r + 2 t \; , $$ so $$ 3r = x+y+z \; \; , \; \; 2s = y - x \; , \; \; \; 6t = 2 z - x - y \; \; . $$ Note $(x,y,z)$ is a rational triple if and only if $(r,s,t)$ is a rational triple. Then $$ x^2 y + y^2 z + z^2 x = 3 r^3 + \left( s^3 + 3 s^2 t - 9 s t^2 - 3 t^3 \right) $$ and $$ 9 \left( x^2 y + y^2 z + z^2 x\right) - (3r)^3 = 9 \left( s^3 + 3 s^2 t - 9 s t^2 - 3 t^3 \right) $$ If, for example, $t \neq 0,$ divide by $t^3$ and we must have a root of $p^3 + 3 p^2 - 9p-3$ which is irreducible. If $s \neq 0$ use the reciprocal. Insisting on rational values, we find that both $s,t$ are zero, so $$ y-x = 0 \; , \; \; \; 2z - x - y = 0 \; , \; $$ and $$ x=y=z $$ Meanwhile, the method answers a simple question, what kind of surface are we describing in $\mathbb R^3 \; ?$ If we have an irrational root $p$ of $p^3 + 3 p^2 - 9p-3=0$ we have some other irrational real $q$ such that $$ 2z-x-y = q(y-x) \; , $$ $$ (q-1)x + (-q-1) y + 2 z = 0 \; \; , $$ which is evidently a plane containing the line $x=y=z.$ I mostly think the surface is three planes, arranged around the line $x=y=z$ at equal angles, like radii of a circle. Indeed, compared with axes given by vectors $v_1 = (-1,1,0)/ \sqrt(2)$ and $v_2 = (-1,-1,2)/ \sqrt(6),$ it seems the three planes are rotated from $v_1$ in the direction of $v_2$ at exactly the three angles $40^\circ, 100^\circ, 160^\circ,$ repetitions at $220^\circ, 280^\circ, 340^\circ,$ so we see every $60^\circ \; .$ As $360/9 = 40$ this has a bit of plausibility.

Next day: confirming the nature of the surface: first, it is defined by the "curve" gotten by intersecting the surface with the plane $x+y+z = 0,$ as it is a "cylinder" over that curve, with axis of translation the expected line $x=y=z.$ If $$ x = X + t \; , \; \; y = Y + t \; , \; \; z = Z + t \; \; , $$ we find $$ (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right) = (X+Y+Z)^3 - 9 \left( X^2 Y + Y^2 Z + Z^2 X \right) $$

Wednesday, finally got it. also Tottenham just scored on Juventus in the Champions League. The roots of $\eta^3 - 3 \eta - 1 = 0$ are $$ A = 2 \cos \left( \frac{7 \pi}{9} \right) \approx -1.532 \; \; \; , B = 2 \cos \left( \frac{5 \pi}{9} \right) \approx -0.347 \; \; \; , C = 2 \cos \left( \frac{ \pi}{9} \right) \approx 1.879 \; \; \; . $$ We get identity $$ \color{red}{ (Ax+By+Cz)(Bx+Cy+Az)(Cx+Ay+Bz) = (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right)} $$ which confirms that the surface is three planes sharing the line $x=y=z,$ as $A+B+C = 0$

Answer 3

First of all $x=y=z$ is a solution to the equality, so we need to prove that all solutions other than this are not valid.

Let $y=a+x$ and $z=b+x$ where $b,a \in Q$

$(x+y+z)^{3} = 9(x^{2}y + y^{2}z + z^{2}x)$

$\Rightarrow (a+b+3x)^{3} = 9(x^{2}(a+x) + (a+x)^{2}(b+x) + (b+x)^{2}x)$

By equating coefficients, we get,

$\Rightarrow (a+b)^{3} = 9(a^{2}b)$

Now let $m= \frac{b}{a}$

$\Rightarrow m^{3} + 3m^{2} - 6m + 1=0$

Let $f(m)= m^{3} + 3m^{2} - 6m + 1$

Now, here is some food for thought,

If we can prove that $b$ or $a$ is irrational , we would arrive at a contradiction because $x,y,z$ are rational numbers.

If $\frac {b}{a}$ is irrational then either $b$ or $a$ will be irrational and we will arrive at a contradiction, which would give us no solution other than $x=y=z$.

So, all we need to do is prove that the cubic $f(m)=0$ has no rational roots.

$m^{3} + 3m^{2} - 6m + 1 = 0$

Let's assume that $f(m)=0$ has rational roots.

Substitute $m=\frac {p}{q}$ , where $gcd(p,q)=1$ and $p,q \in I$

$(\frac {p}{q})^{3} + 3(\frac {p}{q})^{2} - 6(\frac {p}{q}) + 1=0$

$\Rightarrow p(p^{2} + 3pq - 6q^{2}) = -q^{3}$

We know that $gcd(p,q)=1 \Rightarrow gcd(p,q^{3})=1$.

So $p$ must divide $-1$

Therefore the possible values of $p$ are $\pm 1$

Similarly we can write ,

$\Rightarrow q(q^{2} - 6pq + 3p^{2}) = -p^{3}$

So $q$ must divide $-1$

Therefore the possible values of $q$ are $\pm 1$

The possible roots for $f(m)=0$ are $\frac{p}{q} = \pm 1$.

But $f(1)=-1$ and $f(-1)=9$ .

Therefore, $f(m)=0$ has no rational roots.

Therefore, $\frac {b}{a}$ is irrational which is a contradiction.

Hence , $x=y=z$ is the only solution.