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Power

Category: crypto

Points: 208

Solves: 19

Statement

RSA is to boring. Raise to the power of x instead.

power.py

nc hax.allesctf.net 1337

Solution

Given server will ask us for a number x and respond with x**x modulo p, where p is some secret prime value. Additionally, we are given a challenge value everytime we are asked to send x. If we manage to send such value, that x**x modulo p = challenge, we will be rewarded with a flag.

Obviously, our first task is to find the value p, which is unknown to us. We can do this using pretty standard method: send some value x such that x**x > p. The value we get as a response will be equal to x minus some multiple of p. This means that x**x - response will be a multiple of p. Calculating this value for several different x and taking GCD of them will result in p.

The harder part will be to actually find an x value such that x**x modulo p = challenge, once we know p. The solution here is very simple and elegant, yet tricky. As we can send arbitrarily large number (not necessarily smaller than p) we can use the fact, that k*p + x modulo p = x for any integer k. Also from Fermat's theorem we know that a ** (p - 1) modulo p = 1. Therefore a ** l(p - 1) modulo p = 1 for any integer l. Therefore if we can find x such that x = l*(p - 1) + 1 and x = k*p + challenge (1), then x ** x modulo p = x ** (l(p - 1) + 1) modulo p = x ** 1 modulo p = k*p + challenge modulo p = challenge and we are done.

To find x satisfying (1) we simply have to solve congruences: x = 1 modulo (p - 1) and x = challenge modulo p, which can be done easily using either chinese remainder theorem or just simple arithmetics.

Flag: ALLES{n0t_100%_elgamal_but_cl0s3}

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