How to Find All Polynomials Satisfying a Specific Condition

When faced with the task of finding all polynomials p(x) with complex coefficients pxinmathbb{C}[x] that satisfy the equation p(x^2)-2x p(x)-2^2, one must analyze the given condition methodically.

Initial Simplification

Firstly, we notice that the given condition can be rewritten as:

p(x-1)-1p(x-1)-12

This simplifies to the condition:

p(x)2-1p(x-1)-12

This simplified condition must hold for all x.

Transformation and Analysis

We introduce a transformation qx p(x-1) which leads us to the simplified condition:

q(x))2q(x))2

Our goal is to find all polynomials q(x) inmathbb{R}[x] that satisfy this condition.

Key Observations and Solutions

1. Zero Polynomial as a Solution:

The zero polynomial qx 0 is a constant solution since it has every complex number as a root. However, this is an annoying edge case that we can assume is not valid.

2. Nonzero Polynomials:

Assuming qx is nonzero and has a finite number of roots, we can compare the leading coefficients on both sides of the equation. The polynomial must be monic, meaning the coefficient of the highest degree term is 1. This ensures that q(x) is a valid polynomial.

3. Root Analysis:

Suppose rinmathbb{C} is a root of qx. Then, we can write:

q(x-r)x-r)h(x)

Plugging this into the condition:

x-r))2-x-)2rh(x)x-2h(x)

This simplifies to:

x-r))2-x-r))2rh(x)

If rneq 0, the left-hand side is divisible by x ± sqrt{r}. At least one of the roots must satisfy this, so the right-hand side should also be divisible by x ± sqrt{r}. Since it cannot divide x-r^2, it must divide hx^2. As hx is linear, it must divide hx and hence q(x).

We can continue this process, forming an infinite sequence of distinct elements rpmsqrt{ r }dots which must all be roots of qx. This is a contradiction since qx has a finite number of roots.

Conclusion:

Hence, qx can only have 0 as a root. Given that q(x) is monic and any monomial x^n satisfies the condition, we conclude:

qxin{0}cup{xn:ninmathbb{N}}

Reverting back to px via qxp(x-1) gives us:

pxin{0}cup{x1-n:ninmathbb{N}}

This completes the solution to the given polynomial problem.