RSA ENCRYPTION

RSA algorithm is used to encrypt data over the Internet by providing public keys for the data to different users.The RSA cryptosystem is the most widely-used public key cryptography algorithm in the world. It can be used to encrypt a message without the need to exchange a secret key separately.

             The algorithm is fully based on mathematics and is fairly simple to understand.

THE ALGORITHM:

•  Two prime numbers p and q are generated.
•  The Modulus N is found such that N=pq (N should be of required key length k)
• phi= (p-1)*(q-1)
• An Integer between 1 and phi is chosen to be the encryption key "e". This should be such that the HCF (e,phi) =1.
• The Secret exponent "d" -> the decryption key is generated such that it follows 1 < d < phi  and that e*d (mod phi) =1   .
• The message to be encrypted is converted to decimal and i represented by "M".
• After encryption the cypher text is called "C". 
• The public key given is (N,e)
• Encryption is done according to the formula

                                 C=(M^e) mod (N)    --------------------------------------(1)

• Decryption is done according to 

                                M=(C^d) mod (N)    ---------------------------------------(2)

•  The encryption is said to be secure if the value of e is greater than 0x10001 or 65537.
• k is usually 1024 , 2048

HOW TO DECRYPT:

1. If only cypher-text C, modulus N and encryption key e is given. And e = 1

          If e=1 then the cypher-text C is the same numerical equivalent of the message cipher. One just needs to convert the Cypher-text back to text. This situation indicates failure of the Cypher and is a strict no-no case.

   2. If only cypher-text C, modulus N and decryption key d is given.
            This situation should only be dealt by the person for whom the message was intended any one else coming across this situation shows a complete failure of the usage of the encryption.

   3. If cypher-text C, modulus N, phi , decryption key d and encryption key e is given. And the numbers are big.
             The problem be solved using a simple python program which uses the equation (2) to solve the question. The code for the problem is given below with large numbers


N =43210326036290106251088810298667915702007744567467854988107937233622821671752930583378558808381197036542903396534337305723496107154050324953536530993744267386008120658910242143992656773360792182698142273761182047011119261779988377408208858109503177413194541524543447827777831409903664749520527375514748678462101261445039900408999429969510786960471887158957679337190091527913340383310225813322177657695256994315048270952315001100548949554323965404643005206732777361386623591544461533797263652780595639288158892122219626720481472616420118039436680647298714768071797941290307081010086729667552569828323354008441335674251

e =65537

c =37209877026138824302301697292319328185824059912506644719041273895972747926698556612194455367172368277268883774102719113194850674012999971337550561061697826458468363574428378679179721672530515881761949297613967812683948622068020382771205609975220407119022966600675469044732891210789248284410739482876937857726026431593283960162298707892143970513804892248370427455318472402011536095802255121284911517882268092785246985981687913310965628793178703498961617628661113277249071490584774764571170891391355080033791745662119139170834529306548644716069025161989103880671580075279656495472299747401794635781847901588156099495017

phiN = 43210326036290106251088810298667915702007744567467854988107937233622821671752930583378558808381197036542903396534337305723496107154050324953536530993744267386008120658910242143992656773360792182698142273761182047011119261779988377408208858109503177413194541524543447827777831409903664749520527375514748678461684001179681025836243272641520046001108394411608315423967170123709569701951917513847951924936347574671212789694222728087028519090194131250999358518159691864169187723815719657688598576820104123456547706995794001020837756695923476582226622677087727699446323459489485731800382017980557710365098279949382831681952

d=36745622940545343875759976434157197886755506358760982257308567037006074391719705315666781200065625116203199598633622026070492775743618607966652393996419663853350085914252797887742782661594880295499227386075929594641556658033207382848075073319786244161193801795105901007640174254798831863226544268004149920625395925259715625248417078457317167458592444532825039828013768181860349705192246622240475711133444054985367520564542488697167606480838294747710396401026533820191299116891186474240520253953309474166963956334430798860084072450328931700881244717326902973061667440442315315084591766881188371668945135391290476758145

m = hex(pow(c, d, N)).rstrip("L") #the third argument is mod with the first raised to the second
print m
 

   The program will give you a insight into how large numbers are used and how python solves them easily.


this are the few types of RSA that can be cracked the rest cannot be cracked easily by anyone . Bruteforcing can take a long time.