54 lines
1.5 KiB
Markdown
54 lines
1.5 KiB
Markdown
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title = "The Cyber Grabs CTF: Unbr34k4bl3 (942)"
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draft = true
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date = 2022-02-02
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tags = ["ctf", "crypto"]
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languages = ["python"]
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layout = "single"
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math = true
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+++
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Crypto challenge Unbr34k4bl3 from the Cyber Grabs CTF.
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<!--more-->
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> No one can break my rsa encryption, prove me wrong !!
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>
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> Flag Format: cybergrabs{}
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>
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> Author: Mritunjya
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>
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> [output.txt] [source.py]
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[output.txt]: ./output.txt
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[source.py]: ./source.py
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Looking at the source code, this challenge looks like a typical RSA challenge at
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first, but there are some important differences to note:
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- $N = pqr$ (line 34). This is a twist but RSA strategies can easily be
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extended to 3 prime components.
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- $p, q \equiv 3 \mod 4$ (line 19). This suggests that the cryptosystem is
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actually a [Rabin cryptosystem][Rabin].
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- We're not given the public keys $e_1$ and $e_2$, but they are related through
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$x$.
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[Rabin]: https://en.wikipedia.org/wiki/Rabin_cryptosystem
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## Finding $e_1$ and $e_2$
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We know that $e_1$ and $e_2$ are related through $x$, which is some even number
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greater than 2, but we're not given any of their real values. We're also given
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through an oddly-named `functor` function that:
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$$
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\begin{aligned}
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1 + e_1 + e_1^2 + \cdots + e_1^x &= 1 + e_2 + e_2^2 \\\
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\frac{1 - e_1^x}{1 - e_1} &= 1 + e_2 + e_2^2
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\end{aligned}
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$$
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Interestingly enough, since $e_1$ and $e_2$ are primes, that means
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I'd like to thank @10, @sahuang, and @thebishop in the Project Sekai discord for
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their help throughout this challenge.
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