Jacques Gélinas on Fri, 05 Oct 2018 17:05:54 +0200 |
[Date Prev] [Date Next] [Thread Prev] [Thread Next] [Date Index] [Thread Index]
Learning with GP: Factorization of F6=2^64+1 - Landry |
Landry, then aged 82, found the factor 274177 of F6 after six months of work [7]. As discussed in [9], his method is probably modified from the one of Fermat [1, p. 367] presented in [Knuth, II.4.5.4]: if N=uv, then (u-v)^2=(u+v)^2-4N, and digits of perfect squares are very special. ------------------------------ Start cut Knuth454C(N=377) ={ my( x=2*sqrtint(N)+1, y=1, r=sqrtint(N)^2-N ); while(r, while(r>0, r-=y; y+=2); if(r==0,break); r+=x;x+=2 ); return((x-y)/2); } F(n) = 2^2^n + 1; Knuth454C() == 13 Knuth454C(F(4)) == 1 Knuth454C(F(5)) == 641 ------------------------------ End paste Landry used this method until 1867 to find the factors of many numbers of the form 2^n-1,2^n+1 [see 6], but noted that it became unusable when two factors were widely separated. He then shortened the "while" loops by predicting the final values of "y" and "x" [8,9]. Unfortunately, he started by looking for a large factor of F6 and only found 274177 when "there was not much left to do". ------------------------------ Features used [4] Operators: +=, -= Functions: sqrtint, break ------------------------------ Links [7] Fortuné Landry, C.R.Acad.Sc. 91(1880)138, Nouv.Corresp.Math. 6(1880)417 [8] Fortuné Landry, Journal de Mathématiques Élémentaires, 5(1881)3-9. [9] H.C. Williams, How was F6 factored ?, Math.Comp. 61(1993)463-474, http://www.ams.org/journals/mcom/1993-61-203/ (References for Landy and Lucas) Jacques Gélinas