Kvant Math Problem 754
The expression
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m14s
Source on kvant.digital
Problem
- Do there exist polynomials $P=P(x,y,z)$, $Q=Q(x,y,z)$, $R=R(x,y,z)$ in variables $x$, $y$, $z$ such that the identity $$(x-y+1)^3P+(y-z-1)^3Q+(z-2x+1)^3R=1?$$ holds?
- The same question for the identity $$(x-y+1)^3P+(y-z-1)^3Q+(z-x+1)^3R=1.$$
P. Gusyatnikov, Yu. V. Nesterenko
All-Union Mathematical Olympiad for School Students (1982, 10th grade)
Exploration
The expression
$$(x-y+1)^3P+(y-z-1)^3Q+(z-2x+1)^3R$$
or its variant with $(z-x+1)^3$ suggests an ideal-theoretic question in the polynomial ring $\mathbb{K}[x,y,z]$, where $\mathbb{K}$ is a field of characteristic $0$, typically $\mathbb{Q}$ or $\mathbb{R}$.
The central issue is whether the constant polynomial $1$ lies in the ideal generated by the three cubic powers. Since taking radicals does not change the zero set, the first diagnostic step is to examine whether the underlying linear forms can vanish simultaneously.
In the first system, the linear relations are
$$x-y+1=0,\quad y-z-1=0,\quad z-2x+1=0.$$
Solving sequentially produces a consistent solution, so a common zero exists. That immediately suggests that the generated ideal cannot be the entire ring, since any polynomial combination would vanish at that point, contradicting the possibility of obtaining $1$.
In the second system,
$$x-y+1=0,\quad y-z-1=0,\quad z-x+1=0,$$
substitution produces a contradiction, hence no common zero exists. This suggests that the radical of the ideal is the unit ideal. The remaining issue is whether cubes already generate the unit ideal without passing to higher powers.
The key step is the relationship between an ideal and its radical: if the radical is the whole ring, then some power of $1$ lies in the ideal, forcing $1$ itself to lie in the ideal.
Problem Understanding
This is a Type A classification problem.
We are asked whether there exist polynomials $P,Q,R$ such that a fixed linear combination of cubes of three linear polynomials equals the constant polynomial $1$, first in a configuration where the linear forms are compatible, and second where they are inconsistent.
The structure indicates an ideal membership question in $\mathbb{K}[x,y,z]$. The decisive criterion is whether the generating polynomials have a common zero in an algebraic closure. In the first case they do, in the second they do not, leading to different outcomes.
The answers are that no such representation exists in the first identity, while it does exist in the second:
$$\boxed{\text{1) No, 2) Yes}}.$$
Proof Architecture
The first lemma establishes that if polynomials have a common zero, then any polynomial combination of them vanishes at that point, so the constant $1$ cannot belong to the generated ideal.
The second lemma computes the solution set of the linear system in the first identity and shows it is nonempty.
The third lemma computes the solution set of the linear system in the second identity and shows it is empty.
The fourth lemma states that the radical of an ideal generated by powers equals the radical of the ideal generated by the base polynomials.
The fifth lemma applies the fact that an ideal in a polynomial ring over an algebraically closed field whose zero set is empty must have radical equal to the whole ring.
The sixth lemma converts the statement that the radical is the whole ring into actual membership of $1$ in the ideal.
The most delicate point is the transition from radical equality to actual membership in the ideal in the second part.
Solution
Let $\mathbb{K}[x,y,z]$ be a polynomial ring over an algebraically closed field $\mathbb{K}$ of characteristic $0$, for instance $\mathbb{C}$.
First identity
Let
$$f_1=x-y+1,\quad f_2=y-z-1,\quad f_3=z-2x+1.$$
Assume polynomials $P,Q,R$ satisfy
$$f_1^3P+f_2^3Q+f_3^3R=1.$$
Suppose a point $(a,b,c)\in\mathbb{K}^3$ satisfies $f_1(a,b,c)=f_2(a,b,c)=f_3(a,b,c)=0$. Substituting into the identity yields
$$0\cdot P(a,b,c)+0\cdot Q(a,b,c)+0\cdot R(a,b,c)=1,$$
which is impossible. Hence existence of such $P,Q,R$ implies that the system $f_1=f_2=f_3=0$ has no solution.
We solve the system explicitly. From $f_1=0$, $x=y-1$. From $f_2=0$, $y=z+1$. Substituting gives $x=z$. Substituting into $f_3=0$ yields
$$z-2z+1=0,$$
so $z=1$, hence $x=1$ and $y=2$. The system has a solution, contradicting the requirement for membership of $1$ in the ideal generated by $f_1^3,f_2^3,f_3^3$.
At the point $(1,2,1)$, each of $f_1^3,f_2^3,f_3^3$ vanishes, so any polynomial combination also vanishes there, while the constant polynomial does not. Therefore no such polynomials $P,Q,R$ exist.
Second identity
Let
$$g_1=x-y+1,\quad g_2=y-z-1,\quad g_3=z-x+1.$$
Assume
$$g_1^3P+g_2^3Q+g_3^3R=1$$
for some polynomials $P,Q,R$.
We first determine the common zeros of $g_1,g_2,g_3$. From $g_1=0$, $x=y-1$. From $g_2=0$, $y=z+1$. Substituting yields $x=z$. Substituting into $g_3=0$ gives $z-x+1=1$, hence $1=0$, which is impossible. Thus the system $g_1=g_2=g_3=0$ has no solution in $\mathbb{K}^3$.
Let $I=(g_1^3,g_2^3,g_3^3)$. The radical satisfies
$$\sqrt{I}=\sqrt{(g_1,g_2,g_3)}.$$
Since $g_1,g_2,g_3$ have no common zero, Hilbert’s Nullstellensatz implies
$$\sqrt{(g_1,g_2,g_3)}=(1),$$
hence $\sqrt{I}=(1)$.
The condition $1\in \sqrt{I}$ means there exists $m\ge 1$ such that $1^m\in I$, which reduces to $1\in I$. Therefore there exist polynomials $P,Q,R$ such that
$$g_1^3P+g_2^3Q+g_3^3R=1.$$
Verification of Key Steps
The first delicate point is the implication from existence of a common zero to impossibility of representing $1$. Evaluating any polynomial identity at a common zero annihilates each term involving generators, forcing the left-hand side to vanish while the right-hand side remains nonzero.
The second delicate point is the computation of radicals. The identity $\sqrt{(h^k)}=\sqrt{(h)}$ holds because vanishing of a power is equivalent to vanishing of the base polynomial.
The third delicate point is the passage from empty zero set to the unit ideal via Nullstellensatz. Over an algebraically closed field, an ideal with empty algebraic set has radical equal to the whole ring, which forces $1$ into the radical, hence into the ideal itself.
Alternative Approaches
An alternative proof uses syzygies and Gröbner bases. In the second identity, a Gröbner basis computation for $(g_1,g_2,g_3)$ yields $1$, and one can then lift this relation to the cubes by explicit reduction. This approach avoids Nullstellensatz but requires constructive elimination steps that are longer than the ideal-theoretic argument.