Unencryptable

Try my luck with factorizing N using FactorDB. It works for special challenge like this where the N is something kinda special, or we just get lucky that FactorDB has done the factorising work for us ;)

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from Crypto.Util.number import bytes_to_long, long_to_bytes

N = 89820998365358013473897522178239129504456795742012047145284663770709932773990122507570315308220128739656230032209252739482850153821841585443253284474483254217510876146854423759901130591536438014306597399390867386257374956301247066160070998068007088716177575177441106230294270738703222381930945708365089958721
c = 0x5233da71cc1dc1c5f21039f51eb51c80657e1af217d563aa25a8104a4e84a42379040ecdfdd5afa191156ccb40b6f188f4ad96c58922428c4c0bc17fd5384456853e139afde40c3f95988879629297f48d0efa6b335716a4c24bfee36f714d34a4e810a9689e93a0af8502528844ae578100b0188a2790518c695c095c9d677b
# From FactorDB
p = 8239835397208516111720362847949425401045672365829937602117480449316694558226622200110057535873802132963548914201468383545676262090246827792522994758916609
q = 10900824353334471830007307529937357926160386461967884446160315218630687793341471079170750548554707926611542019859296605188535413447791710067186432371970369

b = 0x7fe8cafec59886e9318830f33747cafd200588406e7c42741859e15994ab62410438991ab5d9fc94f386219e3c27d6ffc73754f791e7b2c565611f8fe5054dd132b8c4f3eadcf1180cd8f2a3cc756b06996f2d5b67c390adcba9d444697b13d12b2badfc3c7d5459df16a047ca25f4d18570cd6fa727aed46394576cfdb56b41
e = 0x10001
phi = (p - 1) * (q - 1)
d = pow(e, -1, phi)
print(long_to_bytes(pow(c, d, N)))

The intended solution relies on the observation about RSA Fixed Point, or messages $m$$m$, for some given $e,n$$e, n$ yields:

or, in other words, the polynomial:

has a solution of $x=m$$x = m$. We are given the RSA fixed point in the problem, and e = 65537. Factoring the polynomial of $x^{65537}-x$$x ^ {65537} - x$ should be trivial. As none of the factors of the polynomial is divisible by $n$$n$, they must provide a factor of $n$$n$ when we find the GCD with $n$$n$. Kudos to epistemologist on Cryptohack for the following solution:

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from Crypto.Util.number import long_to_bytes

DATA = 0x372f0e88f6f7189da7c06ed49e87e0664b988ecbee583586dfd1c6af99bf20345ae7442012c6807b3493d8936f5b48e553f614754deb3da6230fa1e16a8d5953a94c886699fc2bf409556264d5dced76a1780a90fd22f3701fdbcb183ddab4046affdc4dc6379090f79f4cd50673b24d0b08458cdbe509d60a4ad88a7b4e2921
N = 0x7fe8cafec59886e9318830f33747cafd200588406e7c42741859e15994ab62410438991ab5d9fc94f386219e3c27d6ffc73754f791e7b2c565611f8fe5054dd132b8c4f3eadcf1180cd8f2a3cc756b06996f2d5b67c390adcba9d444697b13d12b2badfc3c7d5459df16a047ca25f4d18570cd6fa727aed46394576cfdb56b41
e = 0x10001
c = 0x5233da71cc1dc1c5f21039f51eb51c80657e1af217d563aa25a8104a4e84a42379040ecdfdd5afa191156ccb40b6f188f4ad96c58922428c4c0bc17fd5384456853e139afde40c3f95988879629297f48d0efa6b335716a4c24bfee36f714d34a4e810a9689e93a0af8502528844ae578100b0188a2790518c695c095c9d677b

# We have that DATA^e = DATA (n)
# i.e. DATA satisfies the polynomial p(x) = x^65537 - x

assert DATA^e % N == DATA

# Note that
# x^65537 - x = x (x - 1) (x + 1) (x^2 + 1) (x^4 + 1) (x^8 + 1) (x^16 + 1) (x^32 + 1) (x^64 + 1) (x^128 + 1) (x^256 + 1) (x^512 + 1) (x^1024 + 1) (x^2048 + 1) (x^4096 + 1) (x^8192 + 1) (x^16384 + 1) (x^32768 + 1)
# Loop through these factors

polys = [x, x-1] + [(x^(2^i)+1) for i in range(16)]
assert prod(polys) == x^65537 - x

for factor in polys:
    potential_factor = int(factor.polynomial(Zmod(N))(x=DATA))
    if gcd(potential_factor, N) != 1:
        p = gcd(potential_factor, N)
        q = N // p
        break

d = pow(e, -1, (p-1)*(q-1))
print(long_to_bytes(pow(c,d,N)))

This challenge is seemingly based on Shor's algorithm, the following solution from SC4R is left for the future me to digest what's going on. Too lazy to understand at the time of writing.

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# The challenge was based on shor's algorithm, albeit a slightly different implementation than the one on wikipedia

from Crypto.Util.number import *
import gmpy

n = 0x7fe8cafec59886e9318830f33747cafd200588406e7c42741859e15994ab62410438991ab5d9fc94f386219e3c27d6ffc73754f791e7b2c565611f8fe5054dd132b8c4f3eadcf1180cd8f2a3cc756b06996f2d5b67c390adcba9d444697b13d12b2badfc3c7d5459df16a047ca25f4d18570cd6fa727aed46394576cfdb56b41
e = 0x10001
c = 0x5233da71cc1dc1c5f21039f51eb51c80657e1af217d563aa25a8104a4e84a42379040ecdfdd5afa191156ccb40b6f188f4ad96c58922428c4c0bc17fd5384456853e139afde40c3f95988879629297f48d0efa6b335716a4c24bfee36f714d34a4e810a9689e93a0af8502528844ae578100b0188a2790518c695c095c9d677b

DATA = bytes.fromhex("372f0e88f6f7189da7c06ed49e87e0664b988ecbee583586dfd1c6af99bf20345ae7442012c6807b3493d8936f5b48e553f614754deb3da6230fa1e16a8d5953a94c886699fc2bf409556264d5dced76a1780a90fd22f3701fdbcb183ddab4046affdc4dc6379090f79f4cd50673b24d0b08458cdbe509d60a4ad88a7b4e2921")
b = bytes_to_long(DATA)

assert (pow(b, e, n) == b) # Given
# Hence, we can conclude that e-1 is the order of b with respect to n
# Therefore, e-1 is the output of the quantum part of shor's algorithm, with b as the chosen base
r = e-1

# r is even, so I tried to proceed with the next step of the classical part
p = gmpy.gcd(pow(b, r//2, n) - 1, n)
q = gmpy.gcd(pow(b, r//2, n) + 1, n)

# However, p and q turned out to be just 1 and n (trivial factors)
# Therefore, we must look for higher roots of 1 (This is the part that wikipedia skips over)
# https://pdfs.semanticscholar.org/eecc/d242c53ffb17c2756768d1e0ed7c8589f12a.pdf
# This paper mainly discusses odd orders but the same principles can be applied to even ones as well

# The following function bruteforces higher powers of unity until we have a non trivial root
def factor(_b,_r,_n):
    for i in range(2,1000):
        if _r%i == 0:
            t = pow(_b,_r//i,_n)
            if gmpy.gcd(t-1,_n) != 1 and gmpy.gcd(t-1,_n) != _n:
                print ("found p and q")
                p = gmpy.gcd(t-1,_n)
                q = _n//gmpy.gcd(t-1,_n)
                break
    return p,q

p, q = factor(b, e-1, n)
# Once we have p and q, the message can be easily decrypted
tot = (p-1)*(q-1)
d = inverse(e, tot)
flag = long_to_bytes(pow(c, d, n)).decode()

print(flag)