Characteristic value of Fibonacci Sequences

[tex] C(a,b) = a^2 + ab -b^2[/tex]

The characteristic value of a Fibonacci sequence is an interesting property.
1) C(a,b) = C(a,a-b)
2) C(a,b) * C(c,d) = C(ac+ad-bd,bc-ad)
3) C(a,b) * C(a,a-b) = C(2a^2-2ab+b^2,2ab-b^2) =(C(a,b))^2

[tex]C(a,b)^n = C(A_{n},-B_{n})[/tex]
[tex]A_{n} = \sum_{i=0}^{n}F_{i-1}nCia^{i}b^{(n-i)}[/tex]
[tex]B_{n} = \sum_{i=0}^{n}F_{i-2}nCia^{i}b^{(n-i)}[/tex]

Opps the last two equations are sums as i goes from 0 to n
[tex]F_i[/tex] ={-1,1,0,1,1,2,3...} with [tex]F_{0}= 0[/tex]
nCi are the binominal coefficients

Forgive me for this question but what is the:
Characteristic value of Fibonacci Sequences

I wasn't a math major when I was in university.

[tex] C(a,b) = a^2 + ab -b^2[/tex]

The characteristic value of a Fibonacci sequence is an interesting property.
1) C(a,b) = C(a,a-b)
2) C(a,b) * C(c,d) = C(ac+ad-bd,bc-ad)
3) C(a,b) * C(a,a-b) = C(2a^2-2ab+b^2,2ab-b^2) =(C(a,b))^2

[tex]C(a,b)^n = C(A_{n},-B_{n})[/tex]
[tex]A_{n}=\sum_{i=0}^{n}F_{i-1}nCia^{i}b^{(n-i)}[/tex]
[tex]B_{n}=\sum_{i=0}^{n}F_{i-2}nCia^{i}b^{(n-i)}[/tex]

Opps the last two equations are sums as i goes from 0 to n
[tex]F_i[/tex] ={-1,1,0,1,1,2,3...} with [tex]F_{0}= 0[/tex]
nCi are the binominal coefficients

I improved upon the last equation so that the Fibonacci series can be other than the standard Fibonacci series.
Let x and y be [tex]F_0[/tex] and [tex]F_1[/tex] respectively and [tex]F_{n}=F_{n-2}+F_{n-1}[/tex]

Then

[tex]C(a,b)^{n} = C(A_{n},B_{n})/C(x,y)[/tex]

[tex]A_{n} = \sum_{i=0}^{n}F_{i}nCia^{n-i}b^{i}[/tex]

[tex]B_{n} = \sum_{i=0}^{n}F_{i+1}nCia^{n-i}b^{i}[/tex]

Now given that [tex]pCi = 0[/tex] mod p(prime) except for [tex]i=0[/tex] and [tex]i=p[/tex]; C(a,b)^(p)=C(a,b) mod p = C(xa +b* F_{p},ya + bF_{p+1}) /C(x,y) would seem to be a new property of Fibonacci series assuming that each of C(a,b), a and b <> 0 mod p.

This property is the next candidate for proof. Eventually proof of the whole might be found.

My new hypothesis
Let N Be odd and define the subscripts A,B,C,D to be the sequential integers from (N-3)/2 to
(N+3)/2. Then the following equivalence holds if and only if N is Prime.

I check this as far as my excel program would allow (up to prime = 53). This may or may not be useful but it is an improvement on Wilson's Theory. The only doubt I have is the "and only if" part but neither part stands proven.

This looks like the Lucas test, so I imagine the first failure will be at n = 705.

2879;2130859]After considering the actual values of the first few terms as computed from my expressions they are indeed the Lucus numbers and the test does appear to be an equivalent statement of the Lucus test. It is not the only test generated by the study of powers of the Characteristic function. The Lucus test is derived from letting a = 1 and b both = x[sic-a and b both = 1 per my post on March 2 at 9am)] and generating powers of C(1,1) by using each the formulas 1-3 as given in my first post in a particula manner. Basically, if you multiply (a,b)*(a,b) to get (A(2),B(2)) by operation 3[sic- operation 2] and continue doing so to get as per my first post you usually end up with A(n) and B(n) no longer being coprime even if a and b are coprime, but if you alternate the equivalent expressions by letting a= A(1),b=B(1) then doing the multiplication operation 3[sic, should be operation 2] of my first post with (a,a-b)*(A(1),B(1)) to get (A(2),B(2)) then use (a,b)*(A(2),B(2)) to get (A(3),B(3) , i.e. by alternating between (a,a-1)[sic -(a,a-b) and (a,b) as the first ordered pair times the pair (A(n),B(n)) you get coprime A(n),B(n) whenever b is coprime to a. You can also let "a" = x+1 and "b" = x although this generates a triangular series of polynominals for similar to Pascal's triangle, the only similarity is that the generated terms for A(n) and B(n) in the relation C(A(n), B(n)) = C(A,B)^n are polynominals in the nth degree where the first and last term coefficients of A are both the Fibonacci number F(n) and the (first and last term coefficients of B sum to F(n) and also that the remaining terms are all divisible by n if n is prime. Since C(2,1) = 5, then 5*F(n)^2 = 5^n mod n if n is prime. This test (F(n)^2 = 1)is weaker than the Lucus test since there are more presudo primes but apparently holds for all primes other than 5 (this time I checked it out up to the 1,800 th prime). About half of the Lucus presudo primes are also presudo primes for this test, so you can't get a perfect test by a combination of these two tests. I tried other combinations for a and b, but the only apparent tests were variations of these two tests.

I have been limiting my investigation so far by chosing various other expresions in the first degree for a and b, and determining A(n) and B(n) as polynominals of degree n by the multiplication operation, relation 3 of my first post using alternative first order expresions (a,b) and (a,a-b) as above and determining the relation between a and b such that the polynominals of degree n give all terms divisible by n if n is prime with the exception of one or both of the first and last terms.

One way to speed the process is to determine the polynoninals for m = 2n by multiplying (A(n),B(n)) with (A(n),A(n)-B(n)) then multiplying these polynominals together using multiple operations 3 [sic operations 2] to produce the result given earlier in my thread with respect to the coefficients all being a combination of Fibonacci numbers and the Binominal coefficients.

To: Ramsey 2879,

You took a different approach than did Lucas, and that, in and of itself,
makes your discovery unique.

Thanks,

Don.

Your test takes n/2 steps, much worse than trial division.

I don't understand the final step in the test; can you give an example, perhaps for n = 101? I get (a, b, c, d) = (1500520536206896083277, -927372692193078999176, 573147844013817084101, 927372692193078999176).