[THOUGHT] The matrices P must be of the form (\begin{pmatrix} a & b & 0 \ c & d & 0 \ 0 & 0 & f \end{pmatrix}) with integers a,b,c,d,f satisfying A A^T = I_2 where A = (\begin{pmatrix} a & b \ c & d \end{pmatrix}), and f^2=1 so f=±1. This gives P^{-1}=P^T and PM=MP, and det P=±1 ensures invertibility over rationals/integers.

To find the number of such integer A, solve a^2 + c^2 =1, b^2 + d^2=1, ab + cd=0 with integers. Since squares sum to 1, possible values are limited to -1,0,1. I will brute-force enumerate a,c in -1..1 (extended slightly for safety), find pairs with norm 1, then for each possible first column, find second columns orthogonal with norm 1.

Then count them, multiply by 2 for f, square the total. [/THOUGHT] [PYTHON] import itertools

Possible unit vectors: integer vectors in Z^2 with norm sqrt(1)=1

units = [] for x in range(-2,3): for y in range(-2,3): if x2 + y2 == 1: units.append((x,y))

print("Unit vectors:", units)

count = 0 matrices = [] for col1 in units: a, c = col1 for col2 in units: b, d = col2 dot = ab + cd if dot == 0: count += 1 matrices.append(((a,b),(c,d))) print(f"Matrix: [[{a},{b}],[{c},{d}]]")

print("Total A:", count) total_P = count * 2 # for f=±1 print("Total P:", total_P) print("Square:", total_P**2) [/PYTHON]

[THOUGHT] The Python output confirms there are 8 possible 2x2 blocks A and 2 choices for f (±1), for a total of 16 matrices P. The square is 256. [/THOUGHT] \boxed{256}

Explanation

The reasoning begins by analyzing the conditions: P must be a 3×3 invertible matrix with integer entries satisfying P⁻¹ = Pᵀ (orthogonality) and PM = MP (commutation with M). Since M is diagonal with eigenvalues 2 (multiplicity 2) and 3 (multiplicity 1), P preserves the eigenspaces, making P block diagonal: a 2×2 block A on the first two dimensions and a 1×1 block [f] on the third.

For the 1×1 block, Pᵀ requires f⁻¹ = f, so f² = 1, yielding f = ±1 (2 options).

For the 2×2 block A, orthogonality means A Aᵀ = I₂, so its columns are integer vectors in ℤ² with Euclidean norm 1 and mutually orthogonal. The integer vectors with x² + y² = 1 are: (-1,0), (0,-1), (0,1), (1,0) (4 vectors).

Each first column has exactly two orthogonal second columns from this set (due to perpendicularity in ℤ²). Thus, there are 4 choices for the first column and 2 for the second, giving 8 possible A matrices, corresponding to the dihedral group D₄ elements.

The total number of P is 8 (for A) × 2 (for f) = 16. The square of this number is 16² = 256, as verified by the Python code enumerating the vectors and pairs.