Find all the polynomials $P(t)$ of one variable that fullfill the following for all real numbers $x$ and $y$:
$P(x^2-y^2) = P(x+y)P(x-y)$.
1. We start with the given functional equation for the polynomial \( P(t) \):
   \[
   P(x^2 - y^2) = P(x + y)P(x - y)
   \]
   for all real numbers \( x \) and \( y \).

2. Notice that \( x^2 - y^2 \) can be factored as:
   \[
   x^2 - y^2 = (x + y)(x - y)
   \]
   This suggests that we can rewrite the functional equation in terms of the product of sums and differences of \( x \) and \( y \).

3. To explore the solutions, let's consider specific values for \( x \) and \( y \). First, set \( y = 0 \):
   \[
   P(x^2) = P(x)P(x)
   \]
   This simplifies to:
   \[
   P(x^2) = P(x)^2
   \]

4. Next, set \( x = 0 \):
   \[
   P(-y^2) = P(y)P(-y)
   \]
   Since \( P(t) \) is a polynomial, it must be even or odd. If \( P(t) \) is even, then \( P(-y) = P(y) \). If \( P(t) \) is odd, then \( P(-y) = -P(y) \).

5. Consider the polynomial \( P(t) \) of degree \( n \). Suppose \( P(t) = a_n t^n + a_{n-1} t^{n-1} + \cdots + a_1 t + a_0 \). From the equation \( P(x^2) = P(x)^2 \), we can infer that the degree of \( P(x^2) \) is \( 2n \) and the degree of \( P(x)^2 \) is \( 2n \). Therefore, the degree of \( P(t) \) must be consistent with this relationship.

6. If \( P(t) \) is a non-zero polynomial, the simplest non-trivial solution is \( P(t) = t^n \). Let's verify this:
   \[
   P(x^2 - y^2) = (x^2 - y^2)^n
   \]
   and
   \[
   P(x + y)P(x - y) = (x + y)^n (x - y)^n = (x^2 - y^2)^n
   \]
   Both sides match, confirming that \( P(t) = t^n \) is a solution.

7. Additionally, the zero polynomial \( P(t) = 0 \) and the constant polynomial \( P(t) = 1 \) also satisfy the given functional equation:
   \[
   P(x^2 - y^2) = 0 = 0 \cdot 0
   \]
   and
   \[
   P(x^2 - y^2) = 1 = 1 \cdot 1
   \]

8. Therefore, the polynomials that satisfy the given functional equation are \( P(t) = 0 \), \( P(t) = 1 \), and \( P(t) = t^n \) for any non-negative integer \( n \).

The final answer is \(\boxed{P(t) = 0, 1, \text{ or } t^n}\).