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!
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.
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.