[THOUGHT] We need to count the number of 3×3 invertible integer matrices P such that:

1. P^{-1} = P^T (i.e., P is orthogonal over the integers? Actually, integer entries and P^T = P^{-1} implies P is an orthogonal matrix with integer entries. Over integers, such matrices are exactly signed permutation matrices, because integer orthogonal matrices must have columns of length 1 with integer entries, so each column is a standard basis vector multiplied by ±1, and columns are orthogonal.)
2. PM = MP, where M = diag(2,2,3).

So P must commute with M. Let's find conditions on P from commutation.

Let P = (p_{ij}). Then PM = MP. Compute (PM){ij} = sum_k p{ik} m_{kj} = p_{i1} m_{1j} + p_{i2} m_{2j} + p_{i3} m_{3j}. Since M is diagonal, m_{kj} = 0 if k≠j, so (PM){ij} = p{ij} m_{jj}. Similarly, (MP){ij} = sum_k m{ik} p_{kj} = m_{ii} p_{ij} because m_{ik}=0 if i≠k. Thus PM = MP implies p_{ij} m_{jj} = m_{ii} p_{ij} for all i,j. That is: p_{ij} (m_{jj} - m_{ii}) = 0.

Given m_{11}=2, m_{22}=2, m_{33}=3. So for i,j:

• If m_{ii} = m_{jj}, then equation is automatically satisfied (0=0).
• If m_{ii} ≠ m_{jj}, then we must have p_{ij}=0.

Thus:

• For i=1,2 and j=3: m_{ii}=2, m_{jj}=3, so p_{i3}=0.
• For i=3 and j=1,2: m_{ii}=3, m_{jj}=2, so p_{3j}=0.

So P must be block diagonal: top-left 2×2 block can be anything (subject to orthogonality), and bottom-right 1×1 block is p_{33}. Also p_{13}=p_{23}=p_{31}=p_{32}=0.

So P has form: [ a b 0 ] [ c d 0 ] [ 0 0 e ] where a,b,c,d,e are integers.

Now condition P^{-1} = P^T. Since P is block diagonal, its transpose is: [ a c 0 ] [ b d 0 ] [ 0 0 e ] and its inverse is also block diagonal with inverse of top 2×2 block and 1/e for bottom.

Condition P^T = P^{-1} means:

1. For top 2×2 block: [a c; b d]^T = [a b; c d] must equal inverse of [a b; c d]. So the 2×2 block must be an orthogonal integer matrix (2×2 orthogonal with integer entries). Similarly, for the 1×1 block: e must satisfy e = 1/e? Actually P^T bottom-right is e, P^{-1} bottom-right is 1/e (since determinant of whole matrix is det(top)*e, but inverse's bottom-right is 1/e because block diagonal). So we need e = 1/e => e^2 = 1 => e = ±1.

But also orthogonality of whole matrix: columns must be orthonormal (since P^T P = I). Let's derive conditions directly.

Let P be as above. Compute P^T P: P^T = [[a, c, 0], [b, d, 0], [0, 0, e]] P = [[a, b, 0], [c, d, 0], [0, 0, e]] Multiply: P^T P = [ a^2 + c^2, a b + c d, 0 ] [ a b + c d, b^2 + d^2, 0 ] [ 0, 0, e^2 ] This must equal identity I_3. Thus: (1) a^2 + c^2 = 1 (2) b^2 + d^2 = 1 (3) a b + c d = 0 (4) e^2 = 1

All entries are integers. So from (1): a,c integers with squares summing to 1 => possibilities: (a,c) = (±1,0) or (0,±1). Similarly (b,d) = (±1,0) or (0,±1). Condition (3): a b + c d = 0.

Also we need P invertible, which is automatic if e=±1 and top 2×2 block is invertible (determinant nonzero). For orthogonal matrices, determinant is ±1, so invertible.

Now count the number of such matrices.

We can enumerate possibilities for top 2×2 block. It's a 2×2 integer orthogonal matrix. Known that there are 8 such matrices: the dihedral group D4 of symmetries of square (rotations and reflections) with entries in {0, ±1}. Indeed, they are: Rotations: identity [[1,0],[0,1]]; 90° [[0,-1],[1,0]]; 180° [[-1,0],[0,-1]]; 270° [[0,1],[-1,0]]. Reflections: across x-axis [[1,0],[0,-1]]; across y-axis [[-1,0],[0,1]]; across line y=x [[0,1],[1,0]]; across line y=-x [[0,-1],[-1,0]].

Check: all satisfy columns orthonormal.

But we also have condition (3) a b + c d = 0 which is automatically satisfied for orthogonal matrices because columns are orthogonal. Indeed, for orthogonal matrix Q, Q^T Q = I, so off-diagonal entry is dot product of column1 and column2 = 0. So any integer orthogonal 2×2 matrix works.

Thus number of choices for top 2×2 block: 8.

For e: 2 choices (±1).

Thus total number of P matrices = 8 * 2 = 16.

But we must also ensure that P commutes with M. We already enforced block diagonal structure, but does any orthogonal 2×2 block commute with diag(2,2)? Let's check: For a 2×2 matrix Q, Q diag(2,2) = diag(2,2) Q? Since diag(2,2) is scalar multiple of identity in the 2×2 subspace (2I_2), it commutes with any 2×2 matrix. Indeed, for any Q, (2I)Q = 2Q = Q(2I). So commutation is automatic. And the bottom e commutes with 3 because it's scalar. So all block diagonal matrices of that form commute with M.

Thus total count = 16.

The problem asks: "Find the square of the number of ... matrices". So we need to compute N^2 where N = number of such matrices. So answer = 16^2 = 256.

But we should verify by enumerating in Python to be safe.

We'll generate all possible integer matrices P with entries in {-1,0,1} that satisfy P^T P = I and block diagonal structure (p13=p23=p31=p32=0). Actually we can generate all 3x3 matrices with entries from {-1,0,1} that satisfy orthogonality and commutation, but we can also generate top 2x2 block from the 8 possibilities and e from ±1.

Let's verify with Python. [/THOUGHT] [PYTHON] import itertools import numpy as np

Generate all 2x2 integer orthogonal matrices (entries -1,0,1)

We'll generate all 2x2 matrices with entries in {-1,0,1} and check orthogonality.

possible_2x2 = [] for a in (-1,0,1): for b in (-1,0,1): for c in (-1,0,1): for d in (-1,0,1): # Check orthogonality: columns have norm 1 and are orthogonal col1_norm = aa + cc col2_norm = bb + dd dot = ab + cd if col1_norm == 1 and col2_norm == 1 and dot == 0: possible_2x2.append([[a,b],[c,d]]) print("Number of 2x2 integer orthogonal matrices:", len(possible_2x2))

Now build 3x3 P matrices

M = np.diag([2,2,3]) solutions = [] for top in possible_2x2: for e in (-1,1): P = np.zeros((3,3), dtype=int) P[0:2,0:2] = top P[2,2] = e # Check P^T P = I if np.array_equal(P.T @ P, np.eye(3, dtype=int)): # Check PM = MP if np.array_equal(P @ M, M @ P): solutions.append(P)

print("Number of solutions:", len(solutions)) print("Square of number:", len(solutions)**2) [/PYTHON]

[THOUGHT] The Python output confirms there are 16 matrices satisfying the conditions. The problem asks for the square of that number, which is 256. [/THOUGHT] \boxed{256}