OFFSET
1,4
COMMENTS
If (s, y) is a solution then so is (s, -y); if s != 0 then (1/s, y/s^2) is also a solution.
The quartic elliptic curve is birationally equivalent to v^2 = u^3 - 5*u^2 + 5*u with s = (2*v-u) / (4*u-5). All solutions can be generated from multiples of (u,v) = P = (1,1) and the two transformations above.
Let (s, y) be a solution, a = 1 + s, b = 1 + 1/s and c = |y/s|. Then the distance between a*exp(3*i*Pi/5) and 1 + b*exp(2*i*Pi/5) is c, with a, b, c all rational. This allows creating a rigid regular pentagon with idealized Meccano strips - see 't Hooft for the solution corresponding to 3P, and the Mathematics Stack Exchange link for the derivation and solution corresponding to 4P.
LINKS
Jeremy Tan, Table of n, a(n) for n = 1..50
Jeremy Tan, Rigid pentagons and rational solutions of s^4+s^3+s^2+s+1=y^2, Mathematics Stack Exchange, Apr 1 2020.
Gerard 't Hooft, Meccano Math I
EXAMPLE
The values of s in solutions (s, y) with |s| <= 1 begin -1, 0, 1/3, -8/11, -35/123, 627/808, -20965/43993, ...
MATHEMATICA
a[1] = -1; a[2] = 0; a[n_] := Module[{x = 4, y = 2, s, xr}, Do[s = (y-1) / (x-1); xr = s^2 - x + 4; {x, y} = {xr, s(x-xr) - y}, n-2]; s = (2y-x) / (4x-5); Numerator[MinimalBy[{s, 1/s}, Abs][[1]]]]; Table[a[k], {k, 20}] (* Jeremy Tan, Nov 15 2021 *)
PROG
(PARI)
a(n) = {
[u, v] = ellmul(ellinit([0, -5, 0, 5, 0]), [1, 1], n);
s = (2*v-u) / (4*u-5);
if(abs(s)>1, s=1/s);
numerator(s)
}
CROSSREFS
KEYWORD
sign,frac
AUTHOR
Jeremy Tan, Nov 30 2020
STATUS
approved