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Algebra Difficulty 8.6 Shortlist Find the answer

Let A\mathcal{A} denote the set of all polynomials in three variables x,y,zx, y, z with integer coefficients. Let B\mathcal{B} denote the subset of A\mathcal{A} formed by all polynomials which can be expressed as
(x+y+z)P(x,y,z)+(xy+yz+zx)Q(x,y,z)+xyzR(x,y,z)\begin{align*} (x + y + z)P(x, y, z) + (xy + yz + zx)Q(x, y, z) + xyzR(x, y, z) \end{align*}
with P,Q,RAP, Q, R \in \mathcal{A}. Find the smallest non-negative integer nn such that xiyjzkBx^i y^j z^k \in \mathcal{B} for all non-negative integers i,j,ki, j, k satisfying i+j+kni + j + k \geq n.

A number or a short expression. Spacing and $ signs are ignored.

Solution

To solve the given problem, we need to find the smallest non-negative integer n n such that any monomial xiyjzk x^i y^j z^k with i+j+kn i + j + k \geq n can be expressed in the form:

(x+y+z)P(x,y,z)+(xy+yz+zx)Q(x,y,z)+xyzR(x,y,z) (x + y + z)P(x, y, z) + (xy + yz + zx)Q(x, y, z) + xyzR(x, y, z)

where P,Q,R P, Q, R are polynomials with integer coefficients.

### Step-by-step Analysis

1. Understanding the Problem:
- The monomial xiyjzk x^i y^j z^k needs to be expressed as a polynomial that results from the specific linear combination given in the problem.
- We need to analyze the degrees that can be formed by (x+y+z)P (x+y+z)P , (xy+yz+zx)Q (xy+yz+zx)Q , and xyzR xyzR .

2. Degrees of Terms:
- The term (x+y+z)P (x + y + z)P contributes degree deg(P)+1 \deg(P) + 1 .
- The term (xy+yz+zx)Q (xy + yz + zx)Q contributes degree deg(Q)+2 \deg(Q) + 2 .
- The term xyzR xyzR contributes degree deg(R)+3 \deg(R) + 3 .

3. Constructing a Basis for High Degrees:
- For xiyjzk x^i y^j z^k with i+j+k i + j + k sufficiently large, study the combinations of terms that can sum to this degree.
- Notice that:
- (x+y+z)xi1yjzk (x+y+z)x^{i-1}y^jz^k produces monomials like xiyjzk x^iy^jz^k , xi1yj+1zk x^{i-1}y^{j+1}z^k , and xi1yjzk+1 x^{i-1}y^jz^{k+1} .
- (xy+yz+zx)xi1yj1zk (xy+yz+zx)x^{i-1}y^{j-1}z^k produces monomials like xiyjzk x^iy^jz^k , xi1yj+1zk+1 x^{i-1}y^{j+1}z^{k+1} , etc.
- xyzxi1yj1zk1 xyzx^{i-1}y^{j-1}z^{k-1} directly gives xiyjzk x^iy^jz^k .

4. **Inferring the Value of n n :**
- Observe that for i+j+k=3 i + j + k = 3 , the simplest monomial expressions such as x3,y3,z3 x^3, y^3, z^3 can't be formed using any combination of the terms, as these require linear alternation terms which can't have degree less than 3.
- Once i+j+k4 i + j + k \geq 4 , every required monomial can be constructed using the given forms by expressing simpler terms and adding higher degree components systematically using P,Q,R P, Q, R .

5. Conclusion:
- The construction of every monomial becomes feasible for i+j+k4 i + j + k \geq 4 . Therefore, the smallest n n for which each monomial in xiyjzk x^i y^j z^k can be expressed in the form of the given polynomial combination is:

4 \boxed{4}

This reasoning shows that once the total degree i+j+k i + j + k reaches 4, xiyjzkB x^i y^j z^k \in \mathcal{B} , validating n=4 n = 4 as the smallest such integer.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.