# Solution to UTCTF 2019 - Jacobi's Chance Encryption

##### Name:

Jacobi’s Chance Encryption

##### Category:

crypto

##### Points:

750

#### Writeup

We got this challenge text and a implementation of a strange crypto system that we needed to break:

Public Key 569581432115411077780908947843367646738369018797567841

Can you decrypt Jacobi’s encryption?

def encrypt(m, pub_key):

```
bin_m = ''.join(format(ord(x), '08b') for x in m)
n, y = pub_key
def encrypt_bit(bit):
x = randint(0, n)
if bit == '1':
return (y * pow(x, 2, n)) % n
return pow(x, 2, n)
return map(encrypt_bit, bin_m)
```

And also the encrypted flag.

Looking at the implementation it does some really strange things. It loops over every bit in the
plaintext and gets a large random number `x`

. Then it checks if the bit is 1 or 0 and encodes
that information as either y * x^{2} or x^{2} in the congruence class of y.

There is also a public key involved, that’s just two primes multiplied together, but it’s more of a red herring.

This is actually a kind of neat algebra problem, we want to know if a number is a was
a square of itself before it got reduced by modulo `n`

or not, and we know that `n`

and the number
are coprime.

As with all math problems, someone have solved this in sage already. The name of function is
`kronecker`

, so all we had to do was to write a small sage program to reverse the crypto function.

```
file = open("flag.enc", "r")
data = file.readline()
data = data.split(",")
pubkey = list(factor(569581432115411077780908947843367646738369018797567841))
i = 0
str = ""
sol = ""
for b in data:
if b == "":
continue
if kronecker(int(b, 16), pubkey[1][0]) == 0:
str += "1"
else:
str += "0"
i += 1
if i % 8 == 0:
sol += chr(int(str, 2))
str = ""
print(sol)
```

Flag was `utflag{did_u_pay_attention_in_number_theory}`

.