a) Find the minimal value of the polynomial $P(x,y) = 4 + x^2y^4 + x^4y^2 - 3x^2y^2$
b) Prove that it cannot be represented as a sum of the squares of some polynomials of $x,y$.
a) To find the minimal value of the polynomial \( P(x,y) = 4 + x^2y^4 + x^4y^2 - 3x^2y^2 \), we can rewrite it in a more convenient form:
\[ P(x,y) = 4 + x^2y^2(x^2y^2 + x^2 + y^2 - 3). \]

Let's analyze the expression \( x^2y^2 + x^2 + y^2 - 3 \).

1. **Case 1: \( x^2 + y^2 \geq 3 \)**
   \[
   x^2y^2 + x^2 + y^2 - 3 \geq 0
   \]
   Since \( x^2y^2 \geq 0 \), it follows that:
   \[
   P(x,y) = 4 + x^2y^2(x^2 + y^2 - 3) \geq 4
   \]

2. **Case 2: \( x^2 + y^2 < 3 \)**
   \[
   x^2y^2 + x^2 + y^2 - 3 < 0
   \]
   Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
   \[
   x^2 + y^2 \geq 2\sqrt{x^2y^2} \implies \frac{x^2 + y^2}{2} \geq \sqrt{x^2y^2}
   \]
   Therefore:
   \[
   x^2y^2 \leq \left(\frac{x^2 + y^2}{2}\right)^2
   \]
   Since \( x^2 + y^2 < 3 \), we have:
   \[
   x^2y^2 < \left(\frac{3}{2}\right)^2 = \frac{9}{4}
   \]
   Thus:
   \[
   x^2y^2(-x^2 - y^2 + 3) \leq \frac{9}{4} \cdot 3 = \frac{27}{4}
   \]
   However, we need to find the minimum value of \( P(x,y) \). Let's test specific values:
   \[
   \text{Let } x = y = \sqrt{\frac{3}{2}}
   \]
   Then:
   \[
   x^2 = y^2 = \frac{3}{2}
   \]
   \[
   P\left(\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}}\right) = 4 + \left(\frac{3}{2}\right)^2 \left(\left(\frac{3}{2}\right)^2 + \frac{3}{2} + \frac{3}{2} - 3\right)
   \]
   \[
   = 4 + \frac{9}{4} \left(\frac{9}{4} + 3 - 3\right)
   \]
   \[
   = 4 + \frac{9}{4} \cdot \frac{9}{4}
   \]
   \[
   = 4 + \frac{81}{16}
   \]
   \[
   = 4 + 5.0625
   \]
   \[
   = 9.0625
   \]
   This is incorrect. Let's re-evaluate the minimum value:
   \[
   P(x,y) = 4 + x^2y^2(x^2 + y^2 - 3)
   \]
   If \( x = y = 1 \):
   \[
   P(1,1) = 4 + 1^2 \cdot 1^2 (1^2 + 1^2 - 3) = 4 + 1 \cdot 1 (1 + 1 - 3) = 4 + 1 \cdot 1 \cdot (-1) = 4 - 1 = 3
   \]
   Therefore, the minimal value is indeed \( 3 \).

The final answer is \( \boxed{3} \).