Proving the Uniqueness of Integer Solutions in Polynomial Equations

In the field of mathematics, particularly in algebra, often we encounter polynomial equations with integer coefficients. One intriguing problem is to show that if the equation (x - a_1 cdots x - a_n - 1 f(x)g(x)) holds for non-constant polynomials (f(x)) and (g(x)) with integer coefficients, then the product (f(a_i)g(a_i) -1) for (i 1, ldots, n).

Let's begin by embedding the given information into our proof step-by-step. Suppose the polynomial (P(x) x - a_1 cdots x - a_n - 1) can be factored as (P(x) f(x)g(x)), where (f(x)) and (g(x)) are non-constant polynomials with integer coefficients. We need to show that (f(a_i)g(a_i) -1) for each root (a_i) of (P(x)).

Step 1: Integer Coefficients Implication

Since the roots (a_i) are integers and the polynomials (f(x)) and (g(x)) have integer coefficients, both (f(a_i)) and (g(a_i)) must be integers. Given that the product (f(a_i)g(a_i) -1), it follows that one of the values must be 1 and the other must be -1. This is because the only integer pairs whose product is -1 are (1, -1) and (-1, 1).

Step 2: Degree and Roots Analysis

Since (f(x)) and (g(x)) are non-constant polynomials, they have at least one degree greater than zero. The degree of the polynomial (P(x) f(x)g(x)) is the sum of the degrees of (f(x)) and (g(x)). Therefore, the degree of (P(x)) is greater than or equal to (n), where (n) is the number of roots counted with multiplicity.

But, because (P(x)) is a polynomial of degree (n-1) (since the constant term -1 is the only non-zero constant term), the degree of (P(x)) is strictly less than (n). This implies that (P(x)) has at most (n-1) roots, which contradicts the fact that (P(x)) has (n) roots given by (a_1, a_2, ldots, a_n).

Step 3: Polynomial Degeneracy

Given the contradiction from the previous step, we must conclude that the polynomial (P(x) f(x)g(x)) is degenerate. That is, it must be the zero polynomial. The zero polynomial is the only polynomial that can have an infinite number of roots, and in this specific case, the only way (P(x)) can be the zero polynomial is if (f(x)g(x) -x^2), which is impossible as we will now explain.

Step 4: Contradiction in Limits

Considering the polynomial (f(x)g(x) -x^2), we need to examine the behavior of the left-hand side (LHS) and the right-hand side (RHS) as (x) tends to infinity. The LHS, (f(x)g(x)), being a polynomial of degree at least 2, will tend to infinity or negative infinity as (x) increases. However, the RHS, (-x^2), tends to negative infinity as (x) increases.

However, the key insight is that as (x) approaches infinity, the polynomial (f(x)g(x)) cannot consistently represent (-x^2) because their asymptotic behaviors are fundamentally different. Specifically, the polynomial (f(x)g(x)) will grow faster than (-x^2) in the limit, leading to a contradiction.

Conclusion

Therefore, the initial assumption that (P(x) f(x)g(x)) holds for non-constant polynomials (f(x)) and (g(x)) with integer coefficients, and the integer roots (a_i), must be incorrect. The only consistent solution under the given constraints is that (P(x)) is the zero polynomial, which contradicts the original form (x - a_1 cdots x - a_n - 1 f(x)g(x)).

This proof highlights the importance of considering the degrees and behaviors of polynomials in algebraic manipulations and the significance of integer coefficient polynomials in determining unique solutions.