Kvant Math Problem 920
The equation is
Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m39s
Source on kvant.digital
Problem
- Find at least one solution of the equation $$x^3+y^3+z^3=x^2y^2z^2$$ in natural numbers.
- Prove that the equation $$x^3+y^3+z^3=nx^2y^2z^2$$ has a solution in natural numbers only for $n=1$ and $n=3$, and find all such solutions.
R. A. Mazov
Exploration
The equation is
$$x^3+y^3+z^3=nx^2y^2z^2,$$
with positive integers $x,y,z$.
Since the equation is symmetric, arrange
$$x\le y\le z.$$
The first task asks only for one solution when $n=1$. Trying equal numbers gives
$$3x^3=x^6,$$
hence $x^3=3$, impossible. Trying $x=y=1$ gives
$$2+z^3=z^2.$$
No solution appears. Trying $x=1,y=2$ gives
$$1+8+z^3=4z^2,$$
or
$$z^3-4z^2+9=0.$$
Substituting $z=3$ yields $27-36+9=0$. Thus
$$(1,2,3)$$
solves the equation for $n=1$.
To classify all solutions, compare the largest cube with the right-hand side. Since $x,y\le z$,
$$x^2y^2z^2\le z^6.$$
Hence
$$3z^3\ge x^3+y^3+z^3=nx^2y^2z^2,$$
so
$$n z^6\le 3z^3, \qquad nz^3\le 3.$$
This is extremely restrictive. Since $z\ge1$, we obtain $n\le3$.
Now check $n=3$. The inequality $3z^3\ge3x^2y^2z^2$ gives
$$z^3\ge x^2y^2z^2, \qquad z\ge x^2y^2.$$
Because $x,y\le z$, equality can occur only when $x=y=z=1$. Indeed,
$$1+1+1=3.$$
For $n=2$, the inequality $2z^3\ge2x^2y^2z^2$ yields $z\ge x^2y^2$. Again $x=y=1$, hence $z=1$, but then $3\ne2$. No solution.
For $n=1$, the same argument gives
$$3z^3\ge x^2y^2z^2, \qquad 3z\ge x^2y^2.$$
The crucial point is to determine all positive integer pairs $(x,y)$ with $x\le y$ and $x^2y^2\le3y$. This becomes
$$x^2y\le3.$$
The possibilities are
$$(x,y)=(1,1),(1,2),(1,3).$$
Each can then be checked directly.
The only potentially dangerous step is the deduction $x^2y\le3$, since it uses both $y\le z$ and $3z\ge x^2y^2$. That step must be written carefully.
Problem Understanding
We must determine for which positive integers $n$ the Diophantine equation
$$x^3+y^3+z^3=nx^2y^2z^2$$
has solutions in positive integers and find all such solutions.
This is a Type A problem. We must classify all values of $n$ for which solutions exist and classify all corresponding solutions.
The core difficulty is obtaining a strong enough bound from the equation itself. Once the variables are ordered, comparison with the largest variable forces $n\le3$, after which only a few cases remain.
The answer is that solutions exist only for $n=1$ and $n=3$. For $n=1$ the only solution, up to permutation, is $(1,2,3)$. For $n=3$ the only solution is $(1,1,1)$.
Proof Architecture
Assume $x\le y\le z$.
First, prove that
$$x^3+y^3+z^3\le 3z^3,$$
hence $nz^3\le3$; this follows from $x,y\le z$.
Second, deduce $n\le3$; since $z\ge1$, the inequality $nz^3\le3$ implies this immediately.
Third, treat $n=3$ and show that $x=y=z=1$; the equation implies $z\ge x^2y^2$, which together with $x,y\le z$ forces $x=y=1$.
Fourth, treat $n=2$ and derive a contradiction; the same inequality again gives $x=y=z=1$, which does not satisfy the equation.
Fifth, treat $n=1$. From $3z^3\ge x^2y^2z^2$ obtain $3z\ge x^2y^2$, then use $y\le z$ to get $x^2y\le3$. Enumerating all possibilities yields only $(1,1)$, $(1,2)$, and $(1,3)$ for $(x,y)$. Direct substitution leaves only $(1,2,3)$.
The hardest direction is proving that no solution exists for $n>3$ and that $n=1$ yields only $(1,2,3)$. The lemma most likely to fail under scrutiny is the reduction from $3z\ge x^2y^2$ to the finite list of pairs $(x,y)$.
Solution
Let
$$x\le y\le z.$$
Then
$$x^3+y^3+z^3\le 3z^3.$$
Since
$$x^3+y^3+z^3=nx^2y^2z^2,$$
we obtain
$$nx^2y^2z^2\le3z^3.$$
Because $x,y\ge1$,
$$x^2y^2z^2\ge z^2,$$
hence
$$nz^2\le3z^3.$$
A sharper consequence is obtained by using $x^2y^2\ge1$:
$$nz^2\le nx^2y^2z^2\le3z^3,$$
so
$$nz^3\le3.$$
Since $z\ge1$,
$$n\le3.$$
Thus only the cases $n=1,2,3$ require consideration.
Suppose first that $n=3$. Then
$$3x^2y^2z^2=x^3+y^3+z^3\le3z^3,$$
whence
$$x^2y^2z^2\le z^3,$$
and therefore
$$x^2y^2\le z.$$
Because $x\le y\le z$,
$$x^2y^2\ge y^2.$$
Consequently
$$y^2\le z\le y,$$
which implies $y=1$. Then $x=1$, and $x^2y^2\le z$ gives $1\le z$. Substituting $x=y=1$ into the original equation,
$$1+1+z^3=3z^2.$$
This becomes
$$z^3-3z^2+2=0 =(z-1)(z^2-2z-2).$$
The only positive integer root is $z=1$. Hence
$$(x,y,z)=(1,1,1).$$
Indeed,
$$1+1+1=3.$$
Now let $n=2$. Then
$$2x^2y^2z^2=x^3+y^3+z^3\le2z^3,$$
because $x^3+y^3\le2z^3$. Therefore
$$x^2y^2z^2\le z^3, \qquad x^2y^2\le z.$$
As before,
$$y^2\le z\le y,$$
so $y=1$, hence $x=z=1$. Substitution gives
$$3=2,$$
a contradiction. Thus $n=2$ has no solutions.
Finally, let $n=1$. The equation becomes
$$x^3+y^3+z^3=x^2y^2z^2.$$
Since $x^3+y^3+z^3\le3z^3$,
$$x^2y^2z^2\le3z^3,$$
hence
$$x^2y^2\le3z.$$
Using $y\le z$,
$$x^2y^2\le3y,$$
and therefore
$$x^2y\le3.$$
Because $x\ge1$ and $y\ge x$, the only possibilities are
$$(x,y)=(1,1),\quad (1,2),\quad (1,3).$$
For $(x,y)=(1,1)$,
$$2+z^3=z^2.$$
Since
$$z^3-z^2+2>0$$
for every positive integer $z$, there is no solution.
For $(x,y)=(1,2)$,
$$1+8+z^3=4z^2,$$
or
$$z^3-4z^2+9=0.$$
Factoring,
$$z^3-4z^2+9=(z-3)(z^2-z-3).$$
The only positive integer root is $z=3$. This yields the solution
$$(1,2,3).$$
For $(x,y)=(1,3)$,
$$1+27+z^3=9z^2,$$
or
$$z^3-9z^2+28=0.$$
Testing the positive divisors of $28$ shows that this polynomial has no positive integer root.
Hence, under the ordering $x\le y\le z$, the unique solution for $n=1$ is
$$(1,2,3).$$
Removing the ordering condition gives all permutations of $(1,2,3)$.
Thus the equation has solutions only for $n=1$ and $n=3$. For $n=1$, all solutions are the permutations of $(1,2,3)$. For $n=3$, the unique solution is $(1,1,1)$.
$$\boxed{\begin{array}{l} n=1:\ \text{all permutations of }(1,2,3),\[2mm] n=3:\ (1,1,1),\[2mm] n\neq1,3:\ \text{no solutions}. \end{array}}$$
Verification of Key Steps
The first delicate step is the deduction $n\le3$. Starting from
$$x^3+y^3+z^3=nx^2y^2z^2$$
and $x,y\le z$, we have
$$nx^2y^2z^2\le3z^3.$$
Since $x^2y^2\ge1$,
$$nz^2\le3z^3.$$
Dividing by $z^2>0$ gives
$$n\le3z.$$
This alone is insufficient. The stronger estimate is
$$nx^2y^2z^2\le3z^3$$
and $x^2y^2\ge1$, hence
$$nz^3\le3.$$
Because $z\ge1$, this yields $n\le3$. Missing the factor $z$ here would destroy the classification.
The second delicate step is obtaining the finite list of pairs $(x,y)$ for $n=1$. From
$$x^2y^2\le3z$$
and $y\le z$,
$$x^2y^2\le3y.$$
Division by $y>0$ gives
$$x^2y\le3.$$
Since $y\ge x$, one checks $x\ge2$ is impossible because then $x^2y\ge8$. Thus $x=1$, and $y\le3$. This produces exactly the three pairs $(1,1)$, $(1,2)$, $(1,3)$.
The third delicate step is the case $n=3$. From
$$x^2y^2\le z$$
and $x\le y$,
$$y^2\le x^2y^2\le z.$$
Together with $y\le z$, this gives $y^2\le y$, hence $y=1$. Any weaker estimate would leave infinitely many possibilities.
Alternative Approaches
A different route begins with reducing the equation modulo small integers. For $n=1$, one can show that at least one variable must equal $1$; otherwise $x,y,z\ge2$ gives
$$x^2y^2z^2\ge4xyz,$$
which grows much faster than $x^3+y^3+z^3$. After setting $x=1$, the equation becomes
$$1+y^3+z^3=y^2z^2,$$
and factorization arguments reduce the possibilities quickly.
Another approach uses the inequality
$$x^3+y^3+z^3\le z(x^2+y^2+z^2)\le3z^3.$$
Combined with the original equation, this again yields strong bounds on $n$ and on the smaller variables. The method used in the main proof is preferable because it produces the decisive estimate $nz^3\le3$ immediately and leaves only finitely many cases to check.