What are the modular arithmetic properties of iterating a polynomial function ?

Iteration if you aren't familiar, is repeated composition of a function with itself. It follows the rules:$$\begin{eqnarray}f^{n}(x)\circ f^{m}(x) =f^{n+m}(x)\\{(f^{m})}^{n}(x)=f^{mn}(x)\end{eqnarray}$$ Admittedly, this is a rephrasing of a previously deleted question I posted. The reason I'm interested is, it's used a lot in primality testing. The Lucas-Lehmer test, can be reduced to iterations of $2x^2+1$ Mod other values of $2x^2+1$. In this form, it can be restated as:$$f^{p-2}(7)\equiv 0\bmod f^{1}(2^{\frac{p-1}{2}})\iff f^{1}(2^{\frac{p-1}{2}})\in\mathbb{P}$$

I'd like some help simplifying some of this to an easier test. EDIT: https://en.m.wikipedia.org/wiki/Iterated_function does have a few relevant things in it's example table.

EDIT2: we have that we can use polynomial remainder theorem , Fermat's little theorem, Chinese remainder theorem, and that $2x^2-1$ has no x term. This last 1 shows that both $f^1(2^{\frac{p-1}{2}})$ and $f^1(f^1(2^{\frac{p-1}{2}})-2^{\frac{p-1}{2}})$ and the relevant arithmetic progressions, are congruent mod the first. This fact is already known, posted at devalco.de by Bernhard Helmes, a mersenneforum user. $2^{\frac{p-1}{2}}=7\cdot2^{\frac{p-7}{2}}+2^{\frac{p-7}{2}}-1$

EDIT3 : and $f^1(2^{\frac{p-1}{2}})=2^p-1,\text{p odd}$ the real difficulty is to learn more in depth about iteration, ( purple math doesn't go that far) or dealing with a general form polynomial for the iterates. I know the use a real valued expression in proof of sufficiency, of the original form. I see where some of the iteration forms come from. But, just can't easily deal with them in modular arithmetic. If I could, I might be able to use them along with other facts I know about mersenne prime exponents, and whittle things down to solutions.

EDIT4: using the fact we are in the case of $a=2$ and $b=0$ of $ax^2+bx+\frac{b^2-2b-8}{4a}$ which has nth iterate of:$$\frac{2\alpha^{2^n}+2\alpha^{-2^n}-b}{2a}$$ where $$\alpha\equiv 2^{p-2}(2ax+b\pm\sqrt{(2ax+b)^2-16})\pmod{2^p-1}$$

using a=2,b=0,x=7,n=p-2 we get the first value congruent to:

$$2^{p-1}(2^{p-2}(28\pm\sqrt{28^2-16}))^{2^{p-2}}+2^{p-1}(2^{p-2}(28\pm\sqrt{28^2-16}))^{-2^{p-2}}\pmod {2^p-1}$$

which is then congruent to:

$$2^{p-1}(2^{p-2}(28\pm\sqrt{768}))^{2^{p-2}}+2^{p-1}(2^{p-2}(28\pm\sqrt{768}))^{-2^{p-2}}\pmod {2^p-1}$$

which is :

$$2^{p-1}(2^{p\cdot2^{p-2}+2^{p-1}}(28\pm 16\cdot\sqrt{3})^{2^{p-2}})+2^{p-1}(2^{-p\cdot2^{p-2}+2^{p-1}}(28\pm 16\cdot\sqrt{3})^{(-1)\cdot 2^{p-2}})\pmod {2^p-1}$$

I'll do (or try) the binomial expansion sums later. okay yes I'll go back and fix my error.

question now highlighted.

  • 1
    $\begingroup$ What is the question? I'd like some help is not a question. $\endgroup$ – Marc van Leeuwen Mar 15 at 15:30
  • $\begingroup$ highlighted in yellow now. $\endgroup$ – Roddy MacPhee Mar 15 at 15:39
  • $\begingroup$ what needs clarification now ? $\endgroup$ – Roddy MacPhee Mar 17 at 20:57
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    $\begingroup$ What are modular arithmetic properties of any function? $\endgroup$ – Marc van Leeuwen Mar 17 at 21:04
  • $\begingroup$ for polynomials, we can apply polynomial remainder theorem, Fermat, Euler, CRT, etc. $\endgroup$ – Roddy MacPhee Mar 17 at 21:07

Question: What are the modular arithmetic properties of iterating a polynomial function?

Answer: Nobody knows.

Oh, sure, there are some examples where a lot can be said, but by and large, bupkes.

Are you familiar with the Pollard rho method of factorization? You start with a number $n$ you are trying to factor, and a seed $a$ and a polynomial $f$ and you evaluate $a,f(a),f(f(a)),f(f(f(a))),\dots$ all modulo $n$. You keep checking (in a clever way) to see whether the difference of two of these numbers has a (non-trivial) common factor with $n$. If it does, you win – you have found a factor of $n$.

If we knew something about "the modular arithmetic properties of iterating a polynomial function," then we'd know what $a$ to take and what $f$ to take to guarantee quick success, but we don't. Sometimes you fail to get a non-trivial factor, and then you just have to try another polynomial – and that one may fail, too.

  • $\begingroup$ guess I was thinking because both forms are on there That something could be said. $\endgroup$ – Roddy MacPhee Mar 27 at 0:59
  • $\begingroup$ we actually know, if we have n coprime to prime p coefficients to start, that at least n coefficients, can be reduced to 1 at iteration p-1. That comes from Fermat. $\endgroup$ – Roddy MacPhee Jul 15 at 23:49
  • $\begingroup$ Not sure I understand. $f(x)=x+1$ has $n=2$ coefficients prime to $p=3$. The 2nd iteration is $x+2$, which has only one coefficient that reduces to $1\bmod3$. $\endgroup$ – Gerry Myerson Jul 15 at 23:58
  • $\begingroup$ well the constant term can be though of as not a coefficient in simple terms. but I take your point. my point is at least one of the terms created from every non constant term, will have the original coefficient raised to the $p-1$ st power. $\endgroup$ – Roddy MacPhee Jul 16 at 0:36

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