Rating:
# Challenge : The Secured channel
Connecting to the server :
We weren't given the modulus N but g,and r were given and very big secret !
taking a look at information.py given :
check this part :
`again_what = len(bin(p)[2:])`
`this_is_great=pow(2,again_what)`
`well=this_is_great-1`
`having_fun=well^p`
Trying with many values of p we notice that p+having_fun=2**k-1 with k is number bits of p !
Very interesting ! so the function returns p and having_fun+31337(which is q)
so p+q=p+having_fun+31337=pow(2,k)-1+31337=2**k+31336
and p*q=n
for solving p we get the quadtratic equation : p**2-Sp+n=0 with S=p+q and n=p*q !
But we are missing n !
Taking a look at the Help option !
It says not like the encryption as option1 but helps us to get missing part , meaning N ! How is that ?
well :
Basing on that we get the n (script getN.py)
So back to our equation : **pow(p,2)-Sp+n=0**
Solving it : (solvp.py)
`from sympy.solvers import solve`
`from sympy import Symbol`
`from sympy import *`
`p=Symbol('p')`
`S=2**1024+31336`
`n=(value found using getN)`
`print (solve(p**2-S*p+n,p))`
And we got p and q !!
Back to the server , we were given N , g , r and the cipher was lot bigger than the n ..
Well After all it was : [Paillier Cryptosystem](https://en.wikipedia.org/wiki/Paillier_cryptosystem)
Compute the plaintext message as: 
with L is : 
Lamda is : lcm(p-1,q-1) and mu=invert(((pow(int(g),int(lamda),int(n**2))-1)//n),n)
Decrypting the cipher (getMessage.py) and we get :
**Well after all its not that secure , well done , here you go ! have your gift : Securinets{Pa1lli3r??_I_Th1nk_y0u_r0ck!!}**
**Flag : Securinets{Pa1lli3r??_I_Th1nk_y0u_r0ck!!}**
