rsa machine===============
The Challenge------------------
This challenge gave 250 points and had 35 solves.
The challenge description said: `nc rsamachine.wpictf.xyz 31337 (or 31338 or 31339)`.Additionally a python script was given: `rsa_machine_public.py`
Analyzing the script, it became clear that we need to send `getflag` with a valid signature.
I noticed in this line: `(signature,) = privkey.sign(param, None)`, that a given string is signed and not the hash of it. 
This is the vulnerability which allows exitential forgery.When having:s<sub>1</sub> and s<sub>2</sub> you can calculate s<sub>3</sub> without the knowloedge of the private key `d`.
s<sub>1</sub> = m<sub>1</sub><sup>d</sup> mod n
s<sub>2</sub> = m<sub>2</sub><sup>d</sup> mod n
s<sub>1</sub>*s<sub>1</sub> = s<sub>3</sub> = (m<sub>1</sub>*m<sub>1</sub>)<sup>d</sup> mod n
Now I we need to find m<sub>1</sub> and m<sub>2</sub> such that m<sub>1</sub>*m<sub>2</sub> = `b'getflag'`This can be done by simply factorizing the integer representation of the `b'getflag'`.`int.from_bytes(b'getflag', byteorder='big')`
Having m<sub>1</sub> and m<sub>2</sub> we can run our `exploit.py` script. 
`WPI{m411e4b1e_cipher5_4re_d4ngerou5}`
Knowing the procedure, this challenge was pretty straight forward.It was a nice example of a textbook vulnerability.