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# Square CTF 2018 - C2: flipping bits

## Description

Disabling C2 requires cracking a RSA message. You have two ciphertexts. The public key is (e1, n).

Fortunately (this time), space rabiation caused some bit flibs and the second ciphertext was encrypted with a faulty public key (e2, n). Can you recover the plaintexts?

```

ct1 = 13981765388145083997703333682243956434148306954774120760845671024723583618341148528952063316653588928138430524040717841543528568326674293677228449651281422762216853098529425814740156575513620513245005576508982103360592761380293006244528169193632346512170599896471850340765607466109228426538780591853882736654

ct2 = 79459949016924442856959059325390894723232586275925931898929445938338123216278271333902062872565058205136627757713051954083968874644581902371182266588247653857616029881453100387797111559677392017415298580136496204898016797180386402171968931958365160589774450964944023720256848731202333789801071962338635072065

e1 = 13

e2 = 15

modulus = 103109065902334620226101162008793963504256027939117020091876799039690801944735604259018655534860183205031069083254290258577291605287053538752280231959857465853228851714786887294961873006234153079187216285516823832102424110934062954272346111907571393964363630079343598511602013316604641904852018969178919051627

You have two captured ciphertexts. The public key is (e1, n). But,

due to a transient bit flip, the second ciphertext was encrypted with a faulty

public key: (e2, n). Recover the plaintexts.

(The algorithm is RSA.)

```

## Solution

We are provided with two encodings of the same plaintext, using the same modulo but different public keys.

Further, we have `gcd(e1, e2) = gcd(15, 13) = 1`. This means that some values `u` and `v`, so that `u*e1 + v*e2 = 1` holds, exist. We can easily find `u` and `v` thanks to the extended euclidian algorithm.

Now we notice that `ct1 = M^e1 (mod n)` and `ct2 = M^e2 (mod n)`. We compute:

```

ct1^u * ct2^v = M^(u*e1) * M^(v*e2) = M^(u*e1 + v*e2) = M

```

Once we have `M`, all what remains to do is to take its binary representation and decode it as an ASCII string.

```

flag-54d3db5c1efcd7afa579c37bcb560ae0

```

## Code

You can find the sage script we used to get the flag [here](./flag.sage.py).

Original writeup (https://github.com/ctf-epfl/writeups/tree/master/square18/c2_flipping_bits).