> For the complete documentation index, see [llms.txt](https://ir0nstone.gitbook.io/notes/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://ir0nstone.gitbook.io/notes/writeups/picogym/cryptography/no-padding-no-problem.md).

# No Padding, No Problem

Oracles can be your best friend, they will decrypt anything, except the flag's ciphertext. How will you break it? Connect with nc mercury.picoctf.net 10333

Upon connecting, we get the values of $$N$$ and $$e$$ as well as the encrypted ciphertext $$c$$ that represents the flag. We then have a decryption oracle, which can decrypt anything except for the flag.

Note that the ciphertext is decrypted as follows:

$$
m \equiv c^d \mod N
$$

If we ask to decrypt $$-c$$ instead, we get

$$
m \equiv (-c)^d \equiv -c^d \mod N
$$

Note the last congruence is because $$d$$ is odd, so $$(-1)^d = -1$$.

This means that if we pass in the negative of $$c$$, we can get the negative of the decryption!

```
N = 64225632402784743608151428388331019007158039700441403609620876723228303996217136829769322251101831115510439457268097599588978823846061420515078072743333076016253031234729517071419809456539618743788851473244412318432363995783182914809195026673348987512316519371501063936603604905070428868194818209957885002651
R = IntegerModRing(N) 
c = R(23961525860638788006091919862301366730415613260613078904461027043559403510831473561860834624403033454974614369313881141911510211211764847671996788759608002057996932820692709010900418723347410147858586280735791816478632919784849715797867137711835451159040091442311708166252069010315360215005284477472628144578)
print(-c)

# send it back, get result
negative_m = R(64225632402784743608151428388331019007158039700441403609620876723228303996217136829769322251101831115510439457268097599588978823846061420515078072743333076016253031234729517071419809456249343713593001433770955700064908110713217165957916949916605267065613204854099704669280835867601177422810391570120236404254)
long_to_bytes(-m)

# picoCTF{m4yb3_Th0se_m3s54g3s_4r3_difurrent_1772735}
```

{% hint style="info" %}
There are other ways to do it too - you could calculate $$2^{65537} \mod N$$ and multiply $$c$$ by that, which would yield you $$2c$$ after decryption, and you'd just need to halve it, [as described in this writeup](https://github.com/Dvd848/CTFs/blob/master/2021_picoCTF/No_Padding_No_Problem.md).
{% endhint %}


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