For a polynomial $P(x)$ with integer coefficients and a prime $p$, if there is no $n \in \mathbb{Z}$ such that $p|P(n)$, we say that polynomial $P$ [i]excludes[/i] $p$. Is there a polynomial with integer coefficients such that having degree of 5, excluding exactly one prime and not having a rational root?
To determine if there exists a polynomial \( P(x) \) with integer coefficients, degree 5, excluding exactly one prime, and not having a rational root, we can analyze the given polynomial \( P(x) = ((2x)^2 + 3)((2x)^3 - 3) \).

1. **Degree and Rational Root Test:**
   - The polynomial \( P(x) = ((2x)^2 + 3)((2x)^3 - 3) \) is a product of two polynomials:
     \[
     (2x)^2 + 3 = 4x^2 + 3
     \]
     \[
     (2x)^3 - 3 = 8x^3 - 3
     \]
   - The degree of \( P(x) \) is \( 2 + 3 = 5 \), which satisfies the degree requirement.
   - To check for rational roots, we use the Rational Root Theorem. For a polynomial with integer coefficients, any rational root \( \frac{p}{q} \) must have \( p \) dividing the constant term and \( q \) dividing the leading coefficient.
     - For \( 4x^2 + 3 \), the constant term is 3 and the leading coefficient is 4. Possible rational roots are \( \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2} \). None of these are roots of \( 4x^2 + 3 \).
     - For \( 8x^3 - 3 \), the constant term is -3 and the leading coefficient is 8. Possible rational roots are \( \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{4}, \pm \frac{3}{4} \). None of these are roots of \( 8x^3 - 3 \).
   - Therefore, \( P(x) \) has no rational roots.

2. **Exclusion of Prime \( p = 2 \):**
   - We need to show that \( P(x) \) excludes exactly one prime, specifically \( p = 2 \).
   - For \( P(x) = ((2x)^2 + 3)((2x)^3 - 3) \):
     - \( (2x)^2 + 3 = 4x^2 + 3 \) is always odd for any integer \( x \), hence not divisible by 2.
     - \( (2x)^3 - 3 = 8x^3 - 3 \) is also always odd for any integer \( x \), hence not divisible by 2.
   - Therefore, \( P(x) \) excludes the prime \( p = 2 \).

3. **Inclusion of Other Primes:**
   - We need to show that for any prime \( p > 2 \), there exists an integer \( n \) such that \( p \mid P(n) \).
   - **Case 1: \( p = 3 \)**
     - For \( x = 3 \), \( P(3) = ((2 \cdot 3)^2 + 3)((2 \cdot 3)^3 - 3) = (6^2 + 3)(6^3 - 3) = (36 + 3)(216 - 3) = 39 \cdot 213 \). Both 39 and 213 are divisible by 3.
   - **Case 2: \( p \equiv 1 \pmod{3} \)**
     - By Quadratic Reciprocity, there exists \( y \) such that \( p \mid y^2 + 3 \). Let \( y \equiv 2x \pmod{p} \). Then \( p \mid (2x)^2 + 3 \).
   - **Case 3: \( p \equiv 2 \pmod{3} \)**
     - By properties of cubic residues, there exists \( z \) such that \( p \mid z^3 - 3 \). Let \( z \equiv 2x \pmod{p} \). Then \( p \mid (2x)^3 - 3 \).

Hence, the polynomial \( P(x) = ((2x)^2 + 3)((2x)^3 - 3) \) works as it excludes exactly one prime (2) and has no rational roots.

The final answer is \( \boxed{ P(x) = ((2x)^2 + 3)((2x)^3 - 3) } \)