Un simple oracle 2 : Chosen Cipher Attack but with a twist

Task

Same as before except, the modulus is hidden from us.

Process

Well this one is the same as before except for the last step we need the modulus and they just hid it away from us.

Let's take a look at the operation done by the oracle :

$$ m \equiv c^d[n] $$

Well, quick and easy maths, let's try to just input -1 because :

$$ \text{ if d is odd then } \ (-1)^d \equiv n-1 [n] \ \text{ else } \ (-1)^d \equiv 1 [n] $$

Well bingo ! Then we just repeat the same thing as in the first part and get our flag using the python terminal :

404CTF{L3_m0dul3_357_t0uj0ur5_7r0uv4bl3}