```
$ unzip psqs1-0dd2921c9fbdb738e51639801f64164dd144d0771011a1dc3d55da6fbcb0fa02.zip
Archive: psqs1-0dd2921c9fbdb738e51639801f64164dd144d0771011a1dc3d55da6fbcb0fa02.zip
[psqs1-0dd2921c9fbdb738e51639801f64164dd144d0771011a1dc3d55da6fbcb0fa02.zip] cipher password:
extracting: cipher
inflating: pub1.pub
inflating: pub2.pub
```
So we have two public keys and one ciphertext. As the challenge name suggests, it has something with factors of public modulus. Let's extract them:
```
$ grep -v -- ----- pub1.pub | tr -d '\n'|base64 -d | openssl asn1parse -inform DER -i -strparse 19
0:d=0 hl=4 l= 522 cons: SEQUENCE

4:d=1 hl=4 l= 513 prim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

521:d=1 hl=2 l= 3 prim: INTEGER :010001

grep -v -- ----- pub2.pub | tr -d '\n'|base64 -d | openssl asn1parse -inform DER -i -strparse 19
0:d=0 hl=4 l= 522 cons: SEQUENCE
4:d=1 hl=4 l= 513 prim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
521:d=1 hl=2 l= 3 prim: INTEGER :010001
```
In both cases the public exponent is `0x10001 = 16**4 + 1`.

```

from math import gcd
n1 = 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

n2 = 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

print(gcd(n1, n2))
```
A ha! It's bigger than one. We can set
```
g = gcd(n1, n2)
p, q, r = n1//g, g, n2//g

e = 0x10001
from gmpy2 import invert
d1 = int(invert(e, (p-1)*(q-1)))
d2 = int(invert(e, (q-1)*(r-1)))

cipher = int.from_bytes(open('cipher', 'rb').read(), 'big')

dec1 = pow(cipher, d1, n1)
print(dec1.to_bytes(1000, 'big'))

# b'\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x02\xe2H4\xd8\x8b\xd1\x9d2l\x07\x94r\x02 \n,\x0e\xc3\xd4\xbc\xbdfu\xf2\xdc\x8e\xff\x1a\xf03\xdd\x164CT\x04\x97R\x0e\ru\xbfuF`u\xcc\r\x8cp\x0b\xdb\xe8V\xd3\x1b\x12.dZ[]\xa2\x99\xfc\x8a\x1a%\x9d\xd3)() _\xefd\xd1\x0b\x12|\xb6o\x80h\xbco\x8cV\xb7\xb6s\x06{t_\xc6\x0ev\xd3\xab\xe21\x1f\xdb\xa7\xbf[\x0e:\x10\x9eX\x85\x1aY\x95\xc4%\xdf%\xc6\xdfr\xa69dJ\xfd\x10|\xa0\x9f\x9d\xe3\x1c\xe1\x14\xc8\x14\x94\x1e7\xb7\xd5M\xae\xe0\x0bg\xb3B\x91\xc4\xd4\xde\xdd\x81t^Scq\xa8`H\xb5\x0e0\xab\x1a\x18c\xf6f\xec\xf9\xae\xa9\xc5\x8e\\\xc6\x0c6\xa2\x05a \x17\xe6\xc8"5`\x8a\x1f_\xeb\x8eet:\x17\n\x93\xeca\x83C\t=\x82\x1fL\x8eX\x88\tUmL\x9d\x1fb\xbf\x1e\'\xe2\x16\xbd\xac\\\xc7X\xe1\x93\xfb\x923\x91\x95l>\x14\xe8\xa1\xc0(W!\xa5\xbfT\xa4\x0f\xf0\xd9\x15\xc7e\xb5D\xc7\xb7.\x87\xb3?f\xceL\x1c.\xddU\xe9\x8c\xd2\xfc\xd3\xc5\xe1\x12nR\x17~J\xe7.\x9f\'\x9b&/\xb2\xfdw\xaa\xb1\xf3\xdb!A\xd6\x1awS\x10"U\xe4\xf6Z\xaf*\xfa\t\x93\xbf\xb7E\xe3T\xfcd\x9fw\x9c@\xe1J\x1d!\x95=\xf9\xbe\xbb\xf0\x1bf"\xf3\x05\xe3,\xb6\xc6\xff\xb4Dh\x1e\xbf\x99\xb3\x9e\xca\xec\x14\t\xb6\xda\xd3\xf4(\xa5\x19l\xab\xb6\x9c|\x084UayXg4\xd0\x874Z\xab\x13\x14\x1cZ\xef\xf9\x0eMa\xfb\xc35\xbf\xfd\xc2|\xc7\xdb\xbb\xf1;Z\xa8-S\xae@\x9c\xe1]\x04>\x03\t\\n6>\x02\xf1g\xdeMBK\x80\xf6E\xe0@\xe6a\xdfVY\xd4\xb9B\x8fp\xb0\xb3\xe6`\xd0\x7f[\x18\x8c53\xc2\x88\xe21\xe3\x12],\xfdG\x19A".\x9d\x0c\xa6\x90\x07\x0f5\xd5\xe6\x91\xa8\x00SECCON{1234567890ABCDEF}\n'

dec2 = pow(cipher, d2, n2)
print(dec2.to_bytes(1000, 'big'))
```

Flag: `SECCON{1234567890ABCDEF}`