Find all polynomials $P(x)$ with integer coefficients such that the polynomial \[ Q(x)=(x^2+6x+10) \cdot P^2(x)-1 \] is the square of a polynomial with integer coefficients.
To find all polynomials \( P(x) \) with integer coefficients such that the polynomial 
\[ Q(x) = (x^2 + 6x + 10) \cdot P^2(x) - 1 \]
is the square of a polynomial with integer coefficients, we start by assuming that 
\[ Q(x) = R^2(x) \]
for some polynomial \( R(x) \) with integer coefficients. This gives us the equation:
\[ (x^2 + 6x + 10) \cdot P^2(x) - 1 = R^2(x). \]

1. **Substitution and Simplification:**
   Let \( x = X - 3 \). Then, we have:
   \[ Q(X - 3) = ((X - 3)^2 + 6(X - 3) + 10) \cdot P^2(X - 3) - 1. \]
   Simplifying the expression inside the parentheses:
   \[ (X - 3)^2 + 6(X - 3) + 10 = X^2 - 6X + 9 + 6X - 18 + 10 = X^2 + 1. \]
   Therefore, we get:
   \[ Q(X - 3) = (X^2 + 1) \cdot P^2(X - 3) - 1. \]
   Since \( Q(X - 3) = R^2(X - 3) \), we have:
   \[ R^2(X - 3) = (X^2 + 1) \cdot P^2(X - 3) - 1. \]

2. **Finding Polynomials \( f \) and \( g \):**
   We need to find polynomials \( f \) and \( g \) with integer coefficients such that:
   \[ (X^2 + 1) f^2(X) = g^2(X) + 1. \]
   This is a form of the Pell's equation \( x^2 + 1 = 2y^2 \). The positive integer solutions of this equation are given by:
   \[ x_n = \frac{1}{2} \left( (1 + \sqrt{2})^{2n-1} + (1 - \sqrt{2})^{2n-1} \right), \]
   \[ y_n = \frac{1}{2\sqrt{2}} \left( (1 + \sqrt{2})^{2n-1} - (1 - \sqrt{2})^{2n-1} \right). \]

3. **Positive Integer Solutions:**
   Since \( f \) and \( g \) are polynomials with integer coefficients, we can assume that \( f \) and \( g \) take only positive integer values for all points greater than some constant \( w \). We can find \( s \) such that for all \( n \geq s \), \( x_n > w \) and \( y_n > w \). Thus, for \( n \geq s \), \( f(x_n) \) and \( g(x_n) \) are positive integers, and:
   \[ g^2(x_n) - 2(y_n \cdot f(x_n))^2 = -1 \]
   for all \( n \geq s \).

4. **Sequences and Degree Analysis:**
   We can find sequences \( \{a_n\}_{n \geq s} \) and \( \{b_n\}_{n \geq s} \) formed by positive integers such that for all \( n \geq s \):
   \[ g(x_n) = x_{a_n}, \]
   \[ y_n \cdot f(x_n) = y_{b_n}. \]
   Let \( k = \deg(g) \) and \( m = \deg(f) \). We can show that the sequence:
   \[ \left\{ (1 + \sqrt{2})^{2a_n - 1 - k(2n - 1)} \right\} \]
   converges, using the fact that \( g(x_n) = x_{a_n} \). Therefore, we can find \( z \) and \( u \) such that for all \( n > z \):
   \[ 2a_n - 1 - k(2n - 1) = u. \]

5. **Expression for \( g \):**
   From the above, we obtain:
   \[ g(x_n) = \frac{1}{2} \left( (1 + \sqrt{2})^{k(2n - 1) + u} + (1 - \sqrt{2})^{k(2n - 1) + u} \right). \]
   For every \( x \) of the form \( (1 + \sqrt{2})^{2n - 1} \), we must have:
   \[ g\left( \frac{x - 1}{2x} \right) = \frac{1}{2} \left( (1 + \sqrt{2})^u \cdot x^k + (1 - \sqrt{2})^u \cdot \left( -\frac{1}{x} \right)^k \right). \]
   Using continuity arguments, we can show that the above identity is true for all \( x \neq 0 \). Since \( g \) has rational coefficients, \( (1 + \sqrt{2})^u \) must be rational, which implies \( u = 0 \).

6. **Final Expression for \( g \):**
   It follows that for all \( x \):
   \[ g\left( \frac{x - 1}{2x} \right) = \frac{1}{2} \left( x^k + \left( -\frac{1}{x} \right)^k \right). \]
   From this, we can find the expression for \( g \) and subsequently the expression for \( f \).

The final answer is \( \boxed{ P(x) = 0 } \).