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Old 2020-09-12, 20:05   #10
Dr Sardonicus
 
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Feb 2017
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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
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