## Google Beginner CTF: ReadySetAction

**SPOILER WARNING: Go play the CTF at
capturetheflag.withgoogle.com first**

The challenge was a short python script with the following code:

from Crypto.Util.number import * flag = b"REDACTED" p = getPrime(1024) q = getPrime(1024) n = p*q m = bytes_to_long(flag) c = pow(m,3,n) print(c) print(n) #154780 ... 6709 #210348 ... 4477 #(i removed both 617 digit long numbers here)

It was immediatly obvious to me that this has to do with RSA cryptography because it multiplies large primes and then does some powmod magic. Because I didn't remember exactly how RSA worked, I quickly read the Wikipedia article again - specifically section 4.1 Attacks against plain RSA

The first item already seems like the solution:

todo10Which would mean that $\sqrt[3]{c}$ is a solution *if* $m$ is small enough.
However

$c={m}^{3}\text{[PARSE ERROR: Undefined("Command("mod")")]}n\text{[PARSE ERROR: NewLine]}$

so the exponentiation could yield a much higher result that we dont get because it is wrapped at $n$. So we can try out how often it wrapped:

$m=\sqrt[3]{c+x*n}\phantom{\rule{1em}{0ex}}x\in \text{[PARSE ERROR: Undefined("Command("N")")]}\text{[PARSE ERROR: NewLine]}$

where $x$ can be brute-forced if $m$ is sufficiently small. Back in python it looks like this:

c = ... n = ... # simple but bad binary search for n-th root def nth_root(c,e): high = 1 while high**e < c: high *= 2 low = 0 mid = 0 last_mid = 0 while mid**e != c: mid = (low + high) // 2 if last_mid == mid: return 0 # i use 0 to indicate a failure if mid**e > c: high = mid else: low = mid last_mid = mid return mid for x in range(100000): m = nth_root(c + n*x, e) # the probability of finding a perfect cube number is low, so any result is what we want if m != 0: print(long_to_bytes(m))

Thats it! With $x=1831$ we get our flag: `CTF{34sy_RS4_1s_e4sy_us3}`

*Article written by metamuffin, text licenced under CC BY-ND 4.0, non-trivial code blocks under GPL-3.0-only except where indicated otherwise*