To determine all polynomials $P(x)$ that satisfy the given condition for all real values of $x$, we need to solve the functional equation:

$16P(x^2) = [P(2x)]^2$

Step 1: Assume the general form of $P(x)$

Let $P(x)$ be a polynomial of degree $n$. Then we can write:

$P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$

Step 2: Analyze the degrees on both sides

1. The left-hand side is $16P(x^2)$. The polynomial $P(x^2)$ will have the form:

$P(x^2) = a_n x^{2n} + a_{n-1} x^{2(n-1)} + \cdots + a_1 x^2 + a_0$

Thus,

$16P(x^2) = 16a_n x^{2n} + 16a_{n-1} x^{2(n-1)} + \cdots + 16a_1 x^2 + 16a_0$

This polynomial has degree $2n$.

2. The right-hand side is $[P(2x)]^2$. The polynomial $P(2x)$ is:

$P(2x) = a_n (2x)^n + a_{n-1} (2x)^{n-1} + \cdots + a_1 (2x) + a_0 = a_n 2^n x^n + a_{n-1} 2^{n-1} x^{n-1} + \cdots + 2a_1 x + a_0$

The square of $P(2x)$ is:

$[P(2x)]^2 = \left(a_n 2^n x^n + \text{lower degree terms}\right)^2 = a_n^2 4^n x^{2n} + \text{lower degree terms}$

This polynomial also has degree $2n$.

Step 3: Equate the leading coefficients

Since the degrees of both sides must match, the leading coefficient of $x^{2n}$ on the left must equal the leading coefficient of $x^{2n}$ on the right. Equating these gives:

$16a_n = a_n^2 4^n$

Step 4: Simplify and solve for $a_n$

Divide both sides by $a_n$ (assuming $a_n \neq 0$):

$16 = a_n 4^n$

Thus,

$a_n = \frac{16}{4^n} = 4^{2-n}$

Step 5: Check specific degrees

1. For $n = 2$:

   $a_2 = 4^{2-2} = 4^0 = 1$

   This suggests $P(x) = x^2$ as a candidate.

   Substituting $P(x) = x^2$ into the original equation:

   $16P(x^2) = 16x^4$ $[P(2x)]^2 = (4x^2)^2 = 16x^4$

   Both sides match, so $P(x) = x^2$ is indeed a solution.

2. For $n > 2$:

   Substituting into $a_n = 4^{2-n}$ gives non-integer coefficients for higher degrees, which are generally not consistent with polynomials having integer coefficients. This suggests that no higher degree polynomial solutions exist.

3. For $n = 1$:

   Substituting into $a_1 = 4^{2-1} = 4$, the polynomial would be $P(x) = 4x + c$, but this does not satisfy the original equation for all $x$.

Step 6: Conclusion

The only polynomial solution is:

$\boxed{P(x) = x^2}$

This is the unique polynomial that satisfies the given functional equation for all real values of $x$.

Would you like any more details or have any questions?

Here are some related questions to consider:

1. What are functional equations and how do they differ from standard algebraic equations?
2. Can you generalize this approach to functional equations involving different polynomial degrees?
3. How can this problem be adapted if the polynomial involves more complex terms, like trigonometric functions?
4. What happens if the coefficients of the polynomial are complex rather than real?
5. How would you solve similar problems where the equation involves polynomials in two variables?

Tip: When dealing with functional equations, always compare the highest degree terms first to understand the general structure of the solution.