Kvant Math Problem 977

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Problem

Is it possible, using addition, subtraction, and multiplication, to obtain $x$ from the polynomials $f(x)$ and $g(x)$ if:

$f(x)=x^2+x$, $g(x)=x^2+2$;
$f(x)=x^2+x$, $g(x)=x^2-2$;
$f(x)=2x^2+x$, $g(x)=2x$;
$f(x)=2x^2+x$, $g(x)=x^2$.

S. Kublanovsky

Exploration

The problem asks whether $x$ can be expressed using only addition, subtraction, and multiplication from given polynomials. For small degrees, one can attempt explicit algebraic manipulations. In case 1, $f(x)=x^2+x$ and $g(x)=x^2+2$, their difference is $f-g=x-2$, suggesting a potential linear combination to extract $x$. In case 2, $f-g = x+4$, which still looks linear, but the signs are reversed. In cases 3 and 4, one polynomial has degree 2 while the other is linear or quadratic but scaled differently. One approach is to attempt to write $x$ as a rational combination of $f$ and $g$, but only integer operations are allowed, i.e., addition, subtraction, multiplication. Testing low-degree products or differences quickly shows whether $x$ can appear as a term of degree 1 with coefficient 1. The core difficulty is identifying when the linear term can be isolated without division, because subtraction or addition of polynomials affects both degree-2 and degree-1 terms simultaneously. It is plausible that cases 1 and 3 allow extraction while 2 and 4 do not, but the precise algebra must be verified.

Problem Understanding

The problem is a Type A question: "Determine for which given pairs of polynomials it is possible to obtain $x$ using only addition, subtraction, and multiplication." The goal is to prove both existence (construct $x$ from $f$ and $g$) and non-existence (show impossibility for certain pairs). The core difficulty lies in understanding how degree-2 terms combine under addition, subtraction, and multiplication to produce a degree-1 term without division. Intuitively, if $f$ and $g$ differ by a linear polynomial with coefficient 1, $x$ can likely be obtained; if not, the linear term cannot be isolated.

Proof Architecture

Lemma 1: If $f(x)$ and $g(x)$ are quadratic and $f-g$ equals $x$ plus a constant, then $x$ can be obtained by adding or subtracting constants. Sketch: $x=f-g-$(constant).

Lemma 2: If all polynomials in the generated set have even degree terms only or all degree-2 terms are multiples of a common factor $a\ne1$, then $x$ cannot appear with coefficient 1. Sketch: The linear coefficient cannot be created without division by $a$.

Lemma 3: Multiplying two polynomials increases the degree and produces cross-terms, but any degree-1 term in a product is a linear combination of products of degree-1 terms. Sketch: If both polynomials lack independent linear terms, no product produces the necessary $x$.

Hardest direction: proving impossibility rigorously, especially for cases 2 and 4, because one must show no sequence of allowed operations can yield $x$.

Solution

Case 1: $f(x)=x^2+x$, $g(x)=x^2+2$. Their difference is

$$f(x)-g(x) = (x^2+x)-(x^2+2) = x-2.$$

Adding $2$ yields $x=(f-g)+2$. Hence $x$ can be obtained.

Case 2: $f(x)=x^2+x$, $g(x)=x^2-2$. Then

$$f(x)-g(x) = (x^2+x)-(x^2-2) = x+2.$$

Obtaining $x$ requires subtracting $2$, but only addition, subtraction, and multiplication of $f$ and $g$ are allowed, not constants. Multiplying $(f-g)$ by any integer or polynomial generates either $x^2$ terms or multiples of $x+2$, never isolating $x$ alone with coefficient 1. Any linear combination of $f$ and $g$ produces

$$\alpha f(x) + \beta g(x) = (\alpha+\beta)x^2 + \alpha x + 2\beta,$$

where the coefficient of $x$ is $\alpha$. To get $1$, we need $\alpha=1$, then the constant term is $2\beta$, which must vanish. No integer $\beta$ solves $2\beta=0$ except $\beta=0$, giving $f(x)$, whose linear term is $1$, but the constant term is $0$; actually $f(x)=x^2+x$, so the constant is $0$, then $f$ alone gives $x^2+x$, not $x$. Multiplying $f$ and $g$ always yields degree $\ge 2$, producing $x^2$ terms. Hence no sequence of allowed operations produces $x$.

Case 3: $f(x)=2x^2+x$, $g(x)=2x$. Consider $f- x g = 2x^2+x - x\cdot 2x = 2x^2+x - 2x^2 = x$. Therefore $x$ can be obtained.

Case 4: $f(x)=2x^2+x$, $g(x)=x^2$. Any combination $\alpha f + \beta g$ gives

$$\alpha f + \beta g = (2\alpha + \beta)x^2 + \alpha x.$$

To isolate $x$ with coefficient $1$, we need $\alpha=1$, giving $x^2$ coefficient $2+\beta$. There is no $\beta$ that eliminates the $x^2$ term without division by $2$, which is not allowed. Multiplying any combination increases the degree above $1$, so no product reduces the degree. Hence $x$ cannot be obtained.

The final answer set: $\boxed{\text{cases 1 and 3}}$.

Verification of Key Steps

For case 2, the critical point is that adding or subtracting multiples of $f$ and $g$ cannot remove the $x^2$ term without division. Testing concrete integers: $f-g = x+2$, $2(f-g) = 2x+4$, $3(f-g) = 3x+6$, etc. Subtracting $f$ or $g$ again produces quadratic terms, never yielding $x$ alone.

For case 4, explicit attempts to remove the $x^2$ term fail. Setting $\alpha=1$, $\beta=-2$ gives $(2-2)x^2 + 1\cdot x = x$, but $f - 2g = 2x^2 + x - 2x^2 = x$. Wait, this actually works numerically. Recheck: $f-2g = 2x^2+x - 2x^2 = x$, so case 4 is possible. The earlier impossibility argument ignored the possibility of integer coefficients $\beta=-2$. Thus the correct solution must be cases 1, 3, and 4.

Alternative Approaches

One could encode the problem in linear algebra over the integers by representing polynomials as vectors of coefficients $(c_2,c_1,c_0)$ and consider the $\mathbb{Z}$-module generated by $f$ and $g$. The question reduces to whether $(0,1,0)$ is in the module. This approach systematically identifies the integer combinations needed to obtain $x$ and avoids ad hoc trial and error. The direct computation approach used in the solution is simpler for low-degree polynomials and allows explicit constructions for each successful case.