0
$\begingroup$

Fermat's Little Theorem tells us that if $\gcd(x,p)=1$ and $p$ is prime, then $x^{p-1} = 1 \pmod p$. Equivalently, if $a(p)=x^{p-1}-1$, then $a(p-1)=0 \pmod p$ if $p$ is prime. When $p > 2$, $a(p-1)$ is reducible and its irreducible factors are cyclotomic polynomials, more specifically, its irreducible factors are polynomials defining a cyclotomic field. Let $q, q_2, q_3,... q_n$ denote all irreducible polynomial factors of $a(p-1)$. When $a(p-1)$ is factored into linear factors $\pmod p$, then all of its factors are distinct, meaning NO square or repeated factors, the factorization is squarefree!

Now the question is, does there exist a sequence $a(p)$ similar to $x^p-1$, such that all of its irreducible polynomial factors define a cyclotomic field, $a(p)$ does NOT contain any repeated factor (factors into distinct polynomials), and the factorization of $a(p-1)$ for any prime $p$ has $p-2$ distinct linear factors? Also, $a(p)$ cannot be of the form $a(p) = (x - a)^p - (y - b)^p$ for any integers $x, y, a, b$.

After repeatedly for days trying to solve this problem, my guess is, unfortunately no other sequence exists. I may be wrong though. Provide a counterexample sequence if so. Thanks for help!

$\endgroup$
1
$\begingroup$

There is indeed a set of equations similar to this. It is truly facinating. I have been using these very equations for nigh on near next to forty years, if not more. And yes, it generalises in a most spectular way.

An isoseries is a form t(n+1) = a.t(n) - t(n-1). When t(0) = 0, t(1)=1, and t(2) is the short-chord (ie the base of a triangle of two edges) of a polygon, the isoseries makes the chords of a polygon.

The selection of members of an isoseries at a stepping of m, makes an isoseries itself, as t(n+m)=a(m)t(n)-t(m-n). It follows directly that a(m) is itself an isoseries in a, in that a(m+1)=a.a(m)-a(m-1), and that a(0)=2, a(1)=a.

The equations break down like the base-form, into 'algebraic roots', the even ones are exactly the equations that the short-chord of a p/2 solve, the odd ones derive from the even ones in the manner of setting x to -x.

The general series corresponding to $b^n-1$ for these, is to start an isomorphic series at $0, \sqrt{n-2}$, with the shortchord $a = \sqrt{n+2}$. Then the equations below will divide the result when the subscript divides the term number, so J4 divides every fourth term.

J4   x                a(2) = 0.0000000
J6   x-1              a(3) = 1.0000000
J3   x+1
J8   x2 - 2           a(4) = 1.41421356
J10  x2-x-1           a(5) = 1.618033988
J5   x2+x-1
J12  x2 - 3           a(6) = 1.73205080757
J14  x3-x2+2x+1       a(7) = 1.801937736
J7   x3+x2+2x-1
J18  x3-3x-1          a(9) = 1.879385241
J9   x3-3x-1    
J16  x4-4x2+2         a(8) = 1.847759065
J20  x4-5x2+5         a(10) = 1.90211303259
J24  x4-4x2+1         a(12) = 1.93185165259

When one sets x=-3, these equations evaluate to the uniqie factors that occur in the fibonacci series.

The behaviour of this series, is that p can divide either a(p-1) or a(p+1), or some divisor of these (in the manner of fermat's little theorm), and there is an even or odd number of these according to gauss's law.

Cyclotomic Numbers

When a is taken to be the shortchord of a polygon, then the span of the isoseries is the span of chords, say Zn. When this set is taken over the set of [1,w], where w^n = -1, as its simplest root, then this gives the cyclotomic numbers CZn.

The case for the heptagon

This web page http://www.os2fan2.com/p7flat.html shows the heptagonal version of the fibonacci series. It is two-dimensional. Primes for which 7 divides p+1 or p-1, have periods dividing p-1. The remainder have periods that divide p²+p+1. For example, 2 has a 7-place period, and 5 has a 31-period, while 3 has a period of 13. 13, on the other hand, has a 12-place period.

The sevenly flat is generated by the iteration to the right, but in the convergant region, it's $\frac{a^xb^y}{-1+a+b}$. The general form is to add two extra terms, where the numerator a,b are replaced by their isomorphs.

$\endgroup$
  • $\begingroup$ I find the heptagonal Fibonacci series interesting enough, like for the cyclotomic, Fibonacci, and Lucas polynomials, are you also able to list the generating polynomials for the heptagonal Fibonacci series? $\endgroup$ – J. Linne Apr 21 '18 at 16:13
  • $\begingroup$ I did find an equation for it years ago. It is (X + X' + X")/(-1+a+b), where X = A^m B^n, A = a, A' = -b/a, and A" = 1/b, while B = b, B' = 1/a. B" = -a/b. Mostly it's driven by matrices, though. The table is constructed by the iterative process at the right. The convergent region is centred on ab² direction. The corresponding enneagonal flat has not been seen yet, although it is known to exist. $\endgroup$ – wendy.krieger Apr 22 '18 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.