## Robert### Challenge
> Oh, Robert, you can always handle everything!`nc 07.cr.yp.toc.tf 10101`
Upon connection, we see```+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++   hi all, all cryptographers know that fast calculation is not easy! ++   In each stage for given integer m, find number n such that:        ++   carmichael_lambda(n) = m, e.g. carmichael_lambda(2021) = 966       +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++| send an integer n such that carmichael_lambda(n) = 52: ```where we can of course assume we will need to pass a certain number of rounds, and the numbers will grow.
### Solution
Early on, we find [this math stackexchange post](https://math.stackexchange.com/questions/41061/what-is-the-inverse-of-the-carmichael-function), where we already make the comment
> looks hard
as, in general, this problem appears to be at least as hard as factoring $m$.We consider the possibility of factoring $m$ and applying a dynamic programming based approach to group the prime factors of $m$ among the prime factors of what would be $n$.In the end, this did not get implemented, as our intermediate attempts at cheesy solutions converge towards a simpler approach that solves the challenge.The first of these cheesy attempts comes from 3m4 -- while setting the basis for further server communication scripts -- where we simply cross our fingers and hope that $m + 1$ is prime, leading to $n = m + 1$ and $\lambda(n) = m$.While this clears up to 6 rounds on numerous occasions, it appears we'd need to either hammer the server really hard, or find something better.Somewhere during this period of running our cheesy script full of hope, dd suggests that we might be in a situations where $m$ is known to be derived from a semiprime originally, i.e. $m = \lambda(pq)$.Alongside this idea, an attempted solution exploiting that property is proposed, that unfortunately has several flaws and doesn't work against the server.
Basing ourselves on this idea, we can however write the dumbest sage script imaginable for this problem:- Let $D$ be the set of divisors of $m$- Enumerate all $(a, b) \in D^2$- If $a + 1$ is prime, $b + 1$ is prime, *and* $\mathsf{lcm}(a, b) = m$: reply with $n = (a + 1)(b + 1)$
Clearly, *if* our assumed property that $m = \lambda(pq)$ holds, and $m$ does not grow too large to enumerate $D^2$, this should give us a correct solution.
Without all too much hope, we run the following sage script (with the `DEBUG` command line argument for pwntools, so that we can observe the flag should it get sent at the end):
```pythonimport osos.environ["PWNLIB_NOTERM"] = "true"
from pwn import remoteio = remote("07.cr.yp.toc.tf", 10101)
io.recvuntil(b"++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++")io.recvuntil(b"++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++")
proof.arithmetic(False)def reverse_lambda(n):    for x in divisors(n):        for y in divisors(n):            if lcm(x, y) == n and is_prime(x + 1) and is_prime(y + 1):                return (x + 1) * (y + 1) try:    while True:        io.recvuntil(b"carmichael_lambda(n) = ")        integer = ZZ(io.recvuntil(b":")[:-1])        print(f"[*] Reversed: {integer} ->", end=" ", flush=True)        rev = reverse_lambda(integer)        print(f"{rev}")        io.sendline(str(rev).encode())except EOFError:    print("EOF")```
Against our initial expectations, we easily clear more than 10 rounds.Slowing down on an occasional $m$ that might have been hard to factor or have a lot of different divisors, the script happily chugs along without the server closing the connection on it, eventually getting the flag after 20 rounds.
##### Flag`CCTF{Carmichael_numbers_are_Fermat_pseudo_primes}`