We are given the code (and the legend that someone was _very_ insistent we should use this code to generate RSA params and it's totally secure):```#!/usr/bin/env python3
from random import getrandbits, randrange
def generate():	IV = 5326453656607417485924926839199132413931736025515611536784196637666048763966950827721156230379191350953055349475955277628710716247141676818146987326676993279104387449130750616051134331541820233198166403132059774303658296195227464112071213601114885150668492425205790070658813071773332779327555516353982732641; seed = 0; temp = [0, 0]; key = 0	while(key != 2):		if key == 0:			seed = getrandbits(1024) | (2 ** 1023 + 1)		seed_ = seed ^ IV; n = seed_ << 1024 | getrandbits(1024); seed = n//seed | 1		while(1):			seed += 2; pi = seed - 1; b = 0; m = pi;			while (m & 1) == 0:				b += 1				m >>= 1			garbage = []; false_positive = 1			for i in range(min(10, seed - 2)):				a = randrange(2, seed)				while a in garbage:					a = randrange(2, seed)				garbage.append(a); z = pow(a, m, seed)				if z == 1 or z == pi:					continue				for r in range(b):					z = (z * z) % seed;					if z == 1:						break					elif z == pi:						false_positive = 0; break				if false_positive:					break			if not false_positive:				break		temp[key] = seed; key += 1	return(temp[0], temp[1])
def egcd(a, b):	if a == 0:		return (b, 0, 1)	else:		g, y, x = egcd(b % a, a)		return (g, x - (b // a) * y, y)
def inverse(a, m):	g, x, y = egcd(a, m)	if g != 1:		raise Exception('modular inverse does not exist')	else:		return x % m
def RSA():	(p, q) = (0, 0)	while(p == q):		(p, q) = generate()	n = p * q	e = 0x10001	d = inverse(e, (p - 1) * (q - 1))
	return (n, e, d)```and the data generated by this code:```n:  31492593980972292127624243962770751972791586997559415119664067126352874455354554396825959492520866585425367347800693358768073355393830644370233119742738920691048174984688259647179867604016694194959178874017481204930806636137689428300906890690499460758884820784037362414109026624345965004388036791265355660637942956892135275111144499770662995342441666665963192321379766221844532877240853120099814348371659810248827790798567539193378235283620829716728221866131016495050641120781696656274043778657558111564011476899021178414456560902915167861941065608670797745473745460370154604158882840935421329890924498705681322154941e:  65537c:  31105943135542175872131854877903463541878591074483146885600982339634188894256348597314413875362761608401326213641601058666720123544299104566877513595233611507482373096711246358592143276357334231127437090663716106190589907211171164566820101003469295773837109380815526746746678580390779170877213166506863176846508933485178812858166031517243792487958635017917996626849408595921536238471169975280762677305759764602707285224588771643832335444552739959904673158661424651074772864245589406308229908379716500604590198490474257870603210120239125184805345188782082520055749851184516545898673495570079185198108523819932428027921```
This is an example of a backdoor in RSA modulus; the modulus looks just fine and is hard to factor by itself, but the factorization becomes very easy for someone knowing a secret constant used in generation. Namely, the first prime p is random, but the second one is generated as the next prime after ((p ^ IV) \* 2\*\*1024 + randbits(1024)) / p, so the moduls n = p\*q = (p ^ IV) \* 2\*\*1024 + randbits(1024) + p\*(some small number due to search of the next prime); the most significant half of bits of n equals p ^ IV plus some small number.```import gmpy2n = ...e = 65537c = ...IV = ...for i in range(10000):	p=((n>>1024)-i)^IV	if n%p == 0:		breakelse:	print("Factorization not found!")
q = n//pd = gmpy2.invert(e, (p-1)*(q-1))print(bytes.fromhex('%x' % pow(c, d, n)))# prints b'IJCTF{kl3pt0_n0th1n6_uP_mY_s133v3_numb3r5}'```