<?xml version="1.0" encoding="UTF-8"?><rss xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:content="http://purl.org/rss/1.0/modules/content/" xmlns:atom="http://www.w3.org/2005/Atom" version="2.0"><channel><title><![CDATA[RSS Feed]]></title><description><![CDATA[RSS Feed]]></description><link>http://direct.ecency.com</link><image><url>http://direct.ecency.com/logo512.png</url><title>RSS Feed</title><link>http://direct.ecency.com</link></image><generator>RSS for Node</generator><lastBuildDate>Thu, 09 Apr 2026 16:00:20 GMT</lastBuildDate><atom:link href="http://direct.ecency.com/created/factorization/rss.xml" rel="self" type="application/rss+xml"/><item><title><![CDATA[Mathematics - Number Theory - Factorization and Euclidean Algorithm]]></title><description><![CDATA[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.]]></description><link>http://direct.ecency.com/hive-163521/@drifter1/mathematics-number-theory-factorization-and-euclidean-algorithm</link><guid isPermaLink="true">http://direct.ecency.com/hive-163521/@drifter1/mathematics-number-theory-factorization-and-euclidean-algorithm</guid><category><![CDATA[hive-163521]]></category><dc:creator><![CDATA[drifter1]]></dc:creator><pubDate>Fri, 19 Aug 2022 08:55:36 GMT</pubDate><enclosure url="https://images.ecency.com/p/GTCx6Xsvc3wXZhyNxEJ5WDGgQ5T4pEYqbSXVdCHTM5FSmyRpT31Y?format=match&amp;mode=fit" length="0" type="false"/></item><item><title><![CDATA[Cryptography: RSA]]></title><description><![CDATA[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]]></description><link>http://direct.ecency.com/cryptography/@icostan/eli5-cryptography-rsa</link><guid isPermaLink="true">http://direct.ecency.com/cryptography/@icostan/eli5-cryptography-rsa</guid><category><![CDATA[cryptography]]></category><dc:creator><![CDATA[icostan]]></dc:creator><pubDate>Sat, 23 Mar 2019 13:30:15 GMT</pubDate></item><item><title><![CDATA[I wanted to discuss about a result I'm studying: Rossella - factorization algorithm in O(log_2)]]></title><description><![CDATA[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]]></description><link>http://direct.ecency.com/cryptography/@logicalworld/i-wanted-to-discuss-about-a-result-i-m-studying-rossella-factorization-algorithm-in-o-log2</link><guid isPermaLink="true">http://direct.ecency.com/cryptography/@logicalworld/i-wanted-to-discuss-about-a-result-i-m-studying-rossella-factorization-algorithm-in-o-log2</guid><category><![CDATA[cryptography]]></category><dc:creator><![CDATA[logicalworld]]></dc:creator><pubDate>Tue, 25 Sep 2018 08:28:18 GMT</pubDate></item><item><title><![CDATA[Conjecture on RSA factorization O(k)]]></title><description><![CDATA[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)]]></description><link>http://direct.ecency.com/cryptography/@logicalworld/conjecture-on-rsa-factorization-o-k</link><guid isPermaLink="true">http://direct.ecency.com/cryptography/@logicalworld/conjecture-on-rsa-factorization-o-k</guid><category><![CDATA[cryptography]]></category><dc:creator><![CDATA[logicalworld]]></dc:creator><pubDate>Tue, 18 Sep 2018 11:57:06 GMT</pubDate></item><item><title><![CDATA[Conjecture on RSA factorization in (2*i+1)*log_ (2 * i + 1) [N]]]></title><description><![CDATA[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]]></description><link>http://direct.ecency.com/cryptography/@logicalworld/conjecture-on-rsa-factorization-in-2-i-1-log-2-i-1-n</link><guid isPermaLink="true">http://direct.ecency.com/cryptography/@logicalworld/conjecture-on-rsa-factorization-in-2-i-1-log-2-i-1-n</guid><category><![CDATA[cryptography]]></category><dc:creator><![CDATA[logicalworld]]></dc:creator><pubDate>Sat, 15 Sep 2018 14:59:15 GMT</pubDate></item><item><title><![CDATA[RSA factorization conjecture]]></title><description><![CDATA[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]]></description><link>http://direct.ecency.com/cryptography/@logicalworld/rsa-factorization-conjecture</link><guid isPermaLink="true">http://direct.ecency.com/cryptography/@logicalworld/rsa-factorization-conjecture</guid><category><![CDATA[cryptography]]></category><dc:creator><![CDATA[logicalworld]]></dc:creator><pubDate>Thu, 13 Sep 2018 10:39:45 GMT</pubDate></item><item><title><![CDATA[RSA GAME OVER (part II)]]></title><description><![CDATA[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]]></description><link>http://direct.ecency.com/cryptography/@logicalworld/rsa-game-over-part-ii</link><guid isPermaLink="true">http://direct.ecency.com/cryptography/@logicalworld/rsa-game-over-part-ii</guid><category><![CDATA[cryptography]]></category><dc:creator><![CDATA[logicalworld]]></dc:creator><pubDate>Wed, 12 Sep 2018 09:19:57 GMT</pubDate></item><item><title><![CDATA[RSA GAME OVER]]></title><description><![CDATA[Pensate alla funzione F(x)=sqrt[2*x2+(x-1)2+(x+1)^2]=y con x numero naturale ed y pari]]></description><link>http://direct.ecency.com/cryptography/@logicalworld/rsa-game-over</link><guid isPermaLink="true">http://direct.ecency.com/cryptography/@logicalworld/rsa-game-over</guid><category><![CDATA[cryptography]]></category><dc:creator><![CDATA[logicalworld]]></dc:creator><pubDate>Tue, 11 Sep 2018 23:39:51 GMT</pubDate></item><item><title><![CDATA[hey guys another breach in RSA]]></title><description><![CDATA[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]]></description><link>http://direct.ecency.com/cryptography/@logicalworld/hey-guys-another-breach-in-rsa</link><guid isPermaLink="true">http://direct.ecency.com/cryptography/@logicalworld/hey-guys-another-breach-in-rsa</guid><category><![CDATA[cryptography]]></category><dc:creator><![CDATA[logicalworld]]></dc:creator><pubDate>Tue, 11 Sep 2018 08:19:54 GMT</pubDate></item><item><title><![CDATA[hey guys did you know that RSA is broken?]]></title><description><![CDATA[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]]></description><link>http://direct.ecency.com/cryptography/@logicalworld/hey-guys-did-you-know-that-rsa-is-broken</link><guid isPermaLink="true">http://direct.ecency.com/cryptography/@logicalworld/hey-guys-did-you-know-that-rsa-is-broken</guid><category><![CDATA[cryptography]]></category><dc:creator><![CDATA[logicalworld]]></dc:creator><pubDate>Tue, 04 Sep 2018 10:52:57 GMT</pubDate></item><item><title><![CDATA[Factoring Quadratic Trinomials - 2]]></title><description><![CDATA[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]]></description><link>http://direct.ecency.com/math/@mathworksheets/factoring-quadratic-trinomials-2</link><guid isPermaLink="true">http://direct.ecency.com/math/@mathworksheets/factoring-quadratic-trinomials-2</guid><category><![CDATA[math]]></category><dc:creator><![CDATA[mathworksheets]]></dc:creator><pubDate>Fri, 21 Apr 2017 22:05:06 GMT</pubDate><enclosure url="https://images.ecency.com/p/21PRtjKRXPQxwsADmVspzsczTD7VJpL3JZcdUq8PJkWGhR82HjBj2wyAnTxpSkQyhqGjdVp3mpKdAfdi3z6DTeWcTBCLHRmqKHHe6QaxQr1euheh7HdFe4dkPDhz18tijGgh4ieq1SwV2RrwsTVdgtW?format=match&amp;mode=fit" length="0" type="false"/></item></channel></rss>