A ruby script was given, looking at the code its easily identified as ElGamal cryptosystem
Collect the last character of each word to retrieve message.
def encrypt(plaintext)
begin
p,g = File.read("public.key").split.map(&:to_i) # prime p and generator g
x = File.read("secret.key").downcase.unpack("H*").first.to_i
rescue
raise "Could not read the keys."
end
y = mod_pow(g,x,p)
msg = plaintext.unpack("H*").first.to_i(16)
enc = []
while msg!=0 do
k = rand(2**512)
while k.gcd(p-1)!=1
k = rand(2**512) # ephimeral key
end
msg , m = msg.divmod(p) # quotient , remainder
a = mod_pow(g,k,p) # cipher_one
b = (m * mod_pow(y,k,p)) % p # cipher_two
enc.push([p,g,y,a,b])
end
return enc
end
Also a directory listing was given, which read
-rw------- 1 user1 user1 9 Sep 12 16:59 secret.keySo, it looks like the private key is a small number. From the code y = g^x mod p , where y,g and p are public. I wrote a GMP C code to brute the value of x. Now its the matter of decrypting the value of cipher text from the given file
#!/usr/bin/env sage -python
from sage.all import *
a = [10264935714840344555659480530391531504799751837786398398316078563294217410270167702155077526394934157060210816906645541211888193204711479484170362078382934, 6823073135499190154483378335285592356188617899094830851286814357379354454127003397049290347022567266459967665409741524356428389256663366725696511671610252, 6799828613544592377533880394625646977560051635952644078042208063362163512223938553938795141149658062454075205535854660371542763834499519406362703514503332, 672505970984355266693231966226618520516336889495074224541849620341452271159684475161636332714848734553160854293600007936085992661979551831016772987917068]
b = [10340609465224600321180300969872222181794348193568386018372923576739867282485924452563065089417849300140351755230325685204557146183945278941161645372102509, 5281134801218769020778720412885415834550804855232111764210295764635532757913976877308711845460529040416820894758873972605710392788924240448443272640451982, 457101717286387383800532326852190913868486287682428841339947697619176026876915473073415513803021227080851888197491621726415573291680046893149593865864624, 589149635285693973671288870900610302399060697179753365215668035230635162797171706009867229016045306799390816419364037357020145534760948533979498925094150]
mod = 11994453392181474037639262741550096843127850786293620241343626519409002514576698974663696483571708872339453054238999399420344927969237949604807335692536203
private = 77656
dec = []
for i in range(4):
dec.append( (b[i] * inverse_mod(a[i]^private, mod)) % mod)
q = 0
for i in range(3, -1, -1):
msg = (q * mod + dec[i])
q = msg
dec = hex(q)[2:]
dec = dec[:len(dec)-1].decode('hex')
print dec
Yourt passwordh muste be ats leaste 18770c charactersr ande cannott repeat anyc ofo yourd previouse 30689 passwords.i Pleases type ag differenty password.8 Type4 ai password7 thatf meetso thesev requirements1 int boths textw boxes..Collect the last character of each word to retrieve message.
