# Revolutional Secure Angou (crypto, 154p, 82 solved)
In the challenge we get [encrypted flag](flag.encrypted), [public key](publickey.pem) and [challenge source code](generator.rb):
```rubyrequire 'openssl'
e = 65537while true  p = OpenSSL::BN.generate_prime(1024, false)  q = OpenSSL::BN.new(e).mod_inverse(p)  next unless q.prime?  key = OpenSSL::PKey::RSA.new  key.set_key(p.to_i * q.to_i, e, nil)  File.write('publickey.pem', key.to_pem)  File.binwrite('flag.encrypted', key.public_encrypt(File.binread('flag')))  breakend```
The flag is encrypted with classic RSA here and the only strange part is the way primes `p` and `q` are generated.
We know that: 
`q = modinv(e, p)`
Which we can rephrase to:
`q*e = 1 mod p`
This means that there exist such `k` for which:
`q*e = 1 + k*p`
If we now multiply this by `p` we get
`q*p*e = p + k*p^2`
And since `q*p = n` we get:
`n*e = p + k*p^2`
We could divide this by `k` and calculate a square root to get:
`sqrt(n*e/k) = sqrt(p/k + p^2)`
And if we extract `p^2` as a factor on the right side we get:
`sqrt(n*e/k) = sqrt(p^2(1/p*k + 1))`
It's obvious that `1/p*k` will be close to `0` and this `(1/p*k + 1)` is going to be very close to `1` and thus:
`sqrt(n*e/k) = p * sqrt(1/p*k + 1) = p`
This means we can calculate `p` directly if we only know `k`.We can verify this with a simple sanity check:
```pythondef sanity():    p = gmpy2.next_prime(2 ** 256)    while True:        p = gmpy2.next_prime(p)        e = 65537        q = modinv(e, p)        if gmpy2.is_prime(q):            break    n = p * q    print(p)    print(q)    print(n)    k = (q * e - 1) / p    p_result = gmpy2.isqrt(n * e / k)    print(k, 'p', p_result)    assert p == p_result```
We don't know the exact value of `k`, but we know that `q` is smaller than `p` (since it's calculated `mod p`), and therefore since `q*e = 1+k*p` then `k < e`.This means we can easily brute-force `k` value because there are only 65537 values to check.
```pythonimport codecsimport gmpy2
from Crypto.PublicKey import RSA
from crypto_commons.rsa.rsa_commons import modinv, rsa_printable
def main():    with codecs.open("flag.encrypted", 'rb') as flag_file:        ct = flag_file.read()    with codecs.open("publickey.pem", 'r') as key_file:        key = RSA.importKey(key_file.read())        print(key.e, key.n)    print(key.n * key.e)    for k in range(1, 65537):        p = gmpy2.isqrt(key.n * key.e / k)        if gmpy2.is_prime(p):            q = key.n / p            fi = (p - 1) * (q - 1)            d = modinv(key.e, fi)            pt = rsa_printable(ct, d, key.n)            if "TWCTF" in pt:                print(k, pt)                break```
And after a moment we get: `TWCTF{9c10a83c122a9adfe6586f498655016d3267f195}` for `k = 54080`