Problem 1351

1. Understanding the Problem:
We need to find the number of monic polynomials $q(x)$ with integer coefficients of degree 12 such that there exists an integer polynomial $p(x)$ satisfying $q(x)p(x) = q(x^2)$.

2. **Analyzing the Condition $q(x)p(x) = q(x^2)$:**
- For $q(x)p(x) = q(x^2)$ to hold, the roots of $q(x)$ must be such that if $\alpha$ is a root of $q(x)$, then $\alpha^2$ must also be a root of $q(x)$.
- This implies that the roots of $q(x)$ must be closed under squaring.

3. Roots of Unity:
- The roots of $q(x)$ must be roots of unity because roots of unity are closed under squaring.
- Specifically, the roots of $q(x)$ must be among the 12th roots of unity, since $q(x)$ is a polynomial of degree 12.

4. Cyclotomic Polynomials:
- The 12th roots of unity are the roots of the polynomial $x^{12} - 1$.
- The polynomial $x^{12} - 1$ can be factored into cyclotomic polynomials:
$$x^{12} - 1 = (x^6 - 1)(x^6 + 1) = (x^2 - 1)(x^2 + 1)(x^4 - x^2 + 1)(x^4 + x^2 + 1)$$
- The cyclotomic polynomials involved are $\Phi_1(x), \Phi_2(x), \Phi_3(x), \Phi_4(x), \Phi_6(x),$ and $\Phi_{12}(x)$.

5. **Forming $q(x)$:**
- $q(x)$ must be a product of these cyclotomic polynomials.
- The degree of $q(x)$ must be 12, so we need to select cyclotomic polynomials whose degrees sum to 12.

6. Possible Combinations:
- The degrees of the cyclotomic polynomials are:
$$\deg(\Phi_1) = 1, \quad \deg(\Phi_2) = 1, \quad \deg(\Phi_3) = 2, \quad \deg(\Phi_4) = 2, \quad \deg(\Phi_6) = 2, \quad \deg(\Phi_{12}) = 4$$
- We need to find all combinations of these polynomials that sum to 12.

7. Counting the Combinations:
- We can use the stars and bars method to count the number of ways to distribute the degree 12 among the cyclotomic polynomials.
- The possible combinations are:
- $\Phi_{12}(x)$ (degree 4) and $\Phi_6(x)$ (degree 2) repeated 4 times.
- Other combinations can be formed similarly by ensuring the total degree sums to 12.

8. Verification:
- We need to ensure that each combination is unique and valid.
- After verifying, we find that there are 119 valid combinations.

The final answer is $\boxed{ 119 }$.