# 1337UP LIVE CTF 2023

## Really Secure Apparently

> Apparently this encryption is "really secure" and I don't need to worry about sharing the ciphertext, or even these values..
>
> n = 689061037339483636851744871564868379980061151991904073814057216873412583484720768694905841053416938972235588548525570270575285633894975913717130070544407480547826227398039831409929129742007101671851757453656032161443946817685708282221883187089692065998793742064551244403369599965441075497085384181772038720949 e = 98161001623245946455371459972270637048947096740867123960987426843075734419854169415217693040603943985614577854750928453684840929755254248201161248375350238628917413291201125030514500977409961838501076015838508082749034318410808298025858181711613372870289482890074072555265382600388541381732534018133370862587
>
> Author: CryptoCat
>
> [`ciphertext`](https://raw.githubusercontent.com/D13David/ctf-writeups/main/1337uplive/crypto/rsa/ciphertext)

Tags: _crypto_

## Solution

We get a setup for a `RSA` challenge. Since `e` is fairly huge we can assume `d` might be small so we can try our luck with [`Wiener's attack`](https://en.wikipedia.org/wiki/Wiener%27s_attack). I used [`this script`](https://github.com/orisano/owiener/blob/master/owiener.py) to retrieve `d`, the rest then is vanilla `RSA` decryption.

```python
from owiener import *
from Crypto.Util.number import *

n = 689061037339483636851744871564868379980061151991904073814057216873412583484720768694905841053416938972235588548525570270575285633894975913717130070544407480547826227398039831409929129742007101671851757453656032161443946817685708282221883187089692065998793742064551244403369599965441075497085384181772038720949

e = 98161001623245946455371459972270637048947096740867123960987426843075734419854169415217693040603943985614577854750928453684840929755254248201161248375350238628917413291201125030514500977409961838501076015838508082749034318410808298025858181711613372870289482890074072555265382600388541381732534018133370862587

d = attack(e, n)
ct = bytes_to_long(open("ciphertext", "rb").read())
c = pow(ct, d, n)
print(long_to_bytes(c))
```

Flag `INTIGRITI{0r_n07_50_53cur3_m4yb3}`

Original writeup (https://github.com/D13David/ctf-writeups/blob/main/1337uplive/crypto/rsa/README.md).