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.

Overview "If you can't explain it simply, you don't understand it well enough" - Einstein RSA (Rivest-Shamir-Adleman) needs no introduction, it is well known and most used public-key cryptosystem

Hello Profs I wanted to discuss about a result I'm studying for the numbers N = 4 * G + 3 = p * q with G even and p = 4 * h + 3 and q = 4 * k + 1 [it can be extended to all other solutions] thank you

Conjecture on RSA factorization O(k) Esempio N=703 (703-3)/4=175 [175-(4n+1)]/3-[175-(4n+1)]/3-(2n+1)]/3=Z(4n+1) , 4[175-(4n+1)]/3+1=(4Z+1)(4n+1)

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

Definizione 1 Ogni numero dispari, non primo, non divisibile per due e per tre si scrive come somma di numeri dispari consecutivi positivi. Dimostrazione Sia N un numero non divisibile per due e per tre

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