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!


1 Answer 1


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.

  • $\begingroup$ 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, 2018 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. $\endgroup$ Apr 22, 2018 at 6:58

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