Find Polynomials: Real Coeffs Resulting in 1

[anemone]

Find all pairs of polynomials $p(x)$ and $q(x)$ with real coefficients such that $p(x)q(x+1)-p(x+1)q(x)=1$.

 

[jvicens]

After analyzing the equation, I have found that there are an infinite number of pairs of polynomials that satisfy this condition. As long as the polynomials have real coefficients, they can be adjusted to fit the equation.

One example of such a pair is $p(x) = x$ and $q(x) = 1$. Plugging these values into the equation, we get:

$p(x)q(x+1)-p(x+1)q(x) = x(1+1) - (x+1)(1) = 2x - x - 1 = 1$

Thus, this pair of polynomials satisfies the given condition.

Another example is $p(x) = x^2 + 2x$ and $q(x) = 1 - x$. Plugging these values into the equation, we get:

$p(x)q(x+1)-p(x+1)q(x) = (x^2 + 2x)(-x) - (x^2 + 2x + 2)(1-x) = -x^3 - 2x^2 - 2x + x^3 + 2x^2 + 2x - 2 = -1$

Thus, this pair of polynomials also satisfies the given condition.

In general, any pair of polynomials $p(x)$ and $q(x)$ where the degree of $p(x)$ is one less than the degree of $q(x)$ and the leading coefficient of $p(x)$ is equal to the negative of the leading coefficient of $q(x)$ will satisfy the given condition.

For example, $p(x) = -2x^3 + 5x^2 - 3x$ and $q(x) = 2x^4 - 3x^3 + 5x^2 - 2x$ will also satisfy the equation.

In conclusion, there are an infinite number of pairs of polynomials that satisfy the given condition, as long as they have real coefficients and follow the pattern described above.