Introduction Hey it's a me again @drifter1! Today we continue with Mathematics, and more specifically the "Number Theory" series, in order to get into the Factorization and Euclidean Algorithm.

Questa congettura è valida per per N=4G+3 (con opportune modifiche è valida anche per N=4G+1 ) Sia N=a*b con b>a allora o N o 4(G-b)+3 o 4(G-2*b)+3 sono divisibili per 3 se N=(3^n)H con H dispari (il

RSA factorization conjecture This conjecture is valid for N = a * b with a in the form a = 4 * G + 3 The part (h-k)/(a-3)=Z is the conjecture the rest is shown Suppose N = 5369 [34+34+24k](k+1)/2=h , k=(a-7)/4

a^2+na=221 , {a^2 +2[[(a-1)/2]2-1]-[(a-2)2+2[[(a-2-1)/2]^2-1]]-34}/6=k , {(k+1)^2 +2[[((k+1)-1)/2]2-1]-[((k+1)-2)2+2*[[((k+1)-2-1)/2]^2-1]]-34}/6=0

Pensate alla funzione F(x)=sqrt[2*x2+(x-1)2+(x+1)^2]=y con x numero naturale ed y pari

hey guys did you know that RSA is broken? mersenneforum.org -> Miscellaneous Math -> Cooperazione per la fattorizzazione read from post number 47 If you can understand it, offer me a coffee

This is the second lessons in the series of factoring quadratic trinomials. This again covers the very basic type of quadratic trinomial of the type   x²  - bx + c, which differ by a negative