# Thread: Congruence matrix of exponential numbers.

1. ## Congruence matrix of exponential numbers.

Fermat's Little Theorem states that an-1 =/= 1 mod n if a =/= n
I've uncovered some empirical evidence to suggest that if a =/= b =/= n then an-2 =/= bn-2 mod n.
Consider the following "mod 5 matrix".

25 35 45 55
15 1 1 1 1 For clarification, 25 represents all the numbers that are congruent to 2 mod 5
25 4 3 1 2 45 represents all the numbers that are congruent to 4 mod 5
35 4 2 1 3 For example 23 == 3 mod 5 or 78 == 3 mod 5
45 1 4 1 4 Note that the next to the next to the last column confirms Fermat's Little Theorem, and the last column
55 0 0 0 0 confirms that a == an mod n

If we construct a similar matrix for mod 7, mod 11 and mod 13, we will observe the same correlations. In addition we will observe that
all of the congruence values in the third from the last column (equivalent to an-2) are unique with no two the same. If you construct a matrix for
an even number or a non-prime number the congruence values are not unique. These observations lead me to believe that:
If a =/= b =/=n then an-2 =/= bn-2 mod n.

I would appreciate any advice or assistance in proving this conjecture.

2. ## Re: Congruence matrix of exponential numbers.

Since I haven't had anyone reply, either it is not an easy problem to solve or I didn't explain it adequately.
Since the software shifted the top line of the mod 5 matrix, (the top line is the exponent line), let me provide a mod 7 matrix for clarity. I have to eliminate the subscripts since the software doesn't allow it's use for replies. I would also like to correct an error that read "the next to the next column" should just read "the next to the last column".

X 2 3 4 5 6 7

1 1 1 1 1 1 1
2 4 1 2 4 1 2
3 2 6 4 5 1 3
4 2 1 4 2 1 4
5 4 6 2 3 1 5
6 1 6 1 6 1 6
7 0 0 0 0 0 0

Note that n-2 = 5, and the column under the exponent 5 (and all exponents congruent to 5 mod 7) has all unique numbers. Note also that the exponent 6 affirms Fermat's Little Theorem, and the exponent 7 affirms that a^n == a mod n. Most important for my conjecture is that the third from the last column contains all unique number congruences, without duplication.