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 2015-09-07, 23:26 #19 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 100101110110112 Posts A few trivial things about Chebyshev polynomials replace all that "mathlove" wrote in pages of scribbles in a simple way. One has to admire how he repeated copy-pasted pages of similar arguments in each of the primus' questions (of which there are scores and the all look the same, only maybe with different values of c, or base). Do these guys ever get tired of typing?! I think not. Everyone gets their kicks of it. mathlove gets a load of stackexchange's "credits" (like PrimeGrid's, but slightly more useful perhaps) for answering the same question 20 times. Hats off to his/her stamina. Ponder these: 1. https://en.wikipedia.org/wiki/Lissajous_curve : $x=A\sin(at+\delta),\quad y=B\sin(bt)$ 2. Lissajous figures where a=1, b=N (with N a natural number) and $\delta=\frac{N-1}{N}\frac{\pi}{2}\$ are Chebyshev polynomials 3. Chebyshev polynomial $T_n(cos \theta) = cos(n \theta)$ (one of the definitions) 4. Using renormalized $C_n(v) = 2 T_n(v/2), \ C_{km}(x) = C_k(C_m(x))$; in particular $C_{k b^n}(x) = C_b(C_b(C_b(...C_b(C_k(x))...)))$ {n times}. So, the whole iterated calculation is a calculation of T_{k*2^n}(a) and comparing it to T_c(a) with some a value like $a = cos \delta$ . Because of periodicity of cos, the result is obvious. In a way, this is a "Fermat" test z^N = z ("mod" N) in complex plane and with irrational complex $z = e^{i \theta}$ with imaginary half of calculations swept under the carpet, projected on real axis only. (A Frobenius PRP test?) primus has been boring people with these trivialities for years now. Search this forum. And always it was "A new conjectured primality test bla-bla-bla" . Only in 2015, he finally settled down for them as PRP tests, which they are; but as such, they are trivial! All you need to realize is the property 4, which makes the purpose of iterations painfully obvious: you are calculating the $C_{N-c}(a)$, because N-c = k*b^n, then compare that the$cos(N {{N-c} \over N} {\pi \over 2}) = cos(N {-c \over N} {\pi \over 2})$. Of course it is!