20200924, 11:15  #45 
"Roman V. Makarchuk"
Aug 2020
Ukraine
42_{8} Posts 
Ok. Once again, very well. Good, good explanation, that explain all and not explain nothing at the same time)) I'm try to explain, feel free to correct me  I'm not mathematician. Only month ago or so I'm read about modular arithmetic))
Just look at this. u=0 in my notation == The test is Fibonacci PRP test. If we write power of matrix as a*c((ca)/(u1))^2==(u1)^p, for u=0 a*c  ((ca)/(1))^2==(1)^p. Let p be uneven, so a*c(ac)^2==1 for u=2 (Lucas PRP))) > a*c(ca)^2==1 Ok? Still no wise math word?)) Remember, that a,c  always integer!!! Plot the implicit curves, WolframAlfa https://www.wolframalpha.com/input/?...+0%2C+10%7D%5D an what we see? Two hyperbolae, and for any a we can easy compute the difference between c values ((5a^2+4)^.5(5a^24)^.5)/2 Easy to see that this value never be an integer for any integer a. I.e. for any a value form Fibonacci test, there not exist an integer value of c for Lucas test for all values of the p  p is not present in the right side of equation. So Fermat PRP never match with Lucas PRP and wise versa. Its elementary and were here is my mistake? Last fiddled with by RMLabrador on 20200924 at 11:26 
20200924, 13:02  #46 
Feb 2017
Nowhere
3802_{10} Posts 
I don't understand what you're asking. You are looking at det(A^{p}), which is (det(A))^{p} = (u1)^{p}.
We have A = [1,1;1,u], A^{n} = [a_{n}(u), b_{n}(u); b_{n}(u), c_{n}(u)] for integer n. The polynomials a, b, c have integer coefficients (easily proved by induction). The formula (c_{n}(u)  a_{n}(u))/(u1) = b_{n}(u) is correct (easily proved by induction). Congratulations. The polynomial identity a_{n}(u)*c_{n}(u)  b_{n}^{2}(u) = (u1)^{n} is then easily shown to be equivalent to the identity I gave earlier, [recall a_{n}(u) + c_{n}(u) = L_{n}, b_{n}(u) = F_{n}, Δ = (u1)^2 + 4] L_{n}^{2}  Δ*F_{n}^{2} = 4*(u1)^{n}. We have, for p prime, u in Z/pZ, A^{p} == A (mod p) if (Δ/p) = +1; and A^{p} == [u+1,0;0,u+1]  A = [u,1;1,1] (mod p) if (Δ/p) = 1. In either case, the determinant of A^{p} is congruent to u1 (mod p) for u in Z/pZ. For u = 0, 2 we have Δ = 5. Other than the case p = 5, we get det(A^{p}) == 1 (mod p) for u = 2, and det(A^{p}) == 1 (mod p) for u = 0. I don't see what the problem is. 
20200924, 13:34  #47 
"Roman V. Makarchuk"
Aug 2020
Ukraine
22_{16} Posts 
)) There is no problem! Please, read my post above, this IS the proof about Fermat and Lucas probable prime do not inteecept or not? Its important, as far as I'm too do not understead, that I'm write understadeble)))
You stated that Fermat and Lucas test are PRP, so all their combination are PRP too. Thats right if use modulo form only, and have no care about existance of factor in this polynomial. Factor inevery their coefficient lead as to symmetry, the ones is key to proof that even in modulo form the correct, nonprp test can be built. I need somehow post on arxive.org, can i do this without invitation? 
20200924, 14:10  #48  
Feb 2017
Nowhere
EDA_{16} Posts 
Quote:
If you mean that (Δ/p) = 1 and (Δ/p) = +1 never occur simultaneously for a given Δ and p > 2, that is trivial. 

20200924, 14:11  #49 
"Roman V. Makarchuk"
Aug 2020
Ukraine
2×17 Posts 
Certainly, I mean, Fermat and Lucas Pseudoprimes do not intercept at any p

20200924, 14:25  #50 
Feb 2017
Nowhere
2×1,901 Posts 

20200924, 14:47  #52 
"Roman V. Makarchuk"
Aug 2020
Ukraine
100010_{2} Posts 
Fibonacci vs Lucas))) Stupid me  100%. I'n even correct the post, in the post, were link with hyperbolae, started as Fibonacci. Please, exuse me.

20200924, 20:10  #54  
6809 > 6502
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Aug 2003
101×103 Posts
21262_{8} Posts 
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20200924, 20:28  #55  
Aug 2006
13462_{8} Posts 
Quote:


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