View Single Post 2020-09-12, 20:05 #10 Dr Sardonicus   Feb 2017 Nowhere 25×7×17 Posts I did notice one unusual situation: The characteristic polynomial is x^2 - (u + 1)*x + (u - 1); the discriminant is u^2 - 2*u + 5. As long as this is nonzero (mod p), there are two eigenvalues, and it is at least not insane to think about diagonalizing the matrix, which might be probative. If p == 1 (mod 4) however, i2 == -1 (mod p) has two solutions. Taking u = 1 + 2*i makes the discriminant of the characteristic polynomial 0, so it has a repeated factor: x^2 - (2 + 2*i)*x + 2*i = (x - (1 + i))^2. In this case, there is only one eigenvalue, 1 + i, of multiplicity 2. The matrix A - (1+i)I (I = 2x2 identity matrix) is (using Pari-GP syntax) N = [-i,1;1,i] which is nilpotent of index 2. Either column [-i;1] or [1;i] of N serves as an eigenvector of A. Thus A = (1 + i)*I + N. Also, I and N commute, so A^p == (1+i)^p*I + N^p (mod p), and N^p is the 2x2 zero matrix. Thus, A^p == (1+i)^p*I == (1 + i)*I (mod p) in this case. Examples: p = 5, i = 2, u = 1 + 2*2 = 0, 1 + i = 3; or i = 3, u = 1 + 2*3 = 2, 1 + i = 4 p = 13, i = 5, u = 1 = 2*5 = 11, 1 + i = 6; or i = 8, u = 1 + 2*8 = 4, 1 + i = 9 Last fiddled with by Dr Sardonicus on 2020-09-12 at 20:08 Reason: xingif ostpy  