Tags: rsa openssl asymmetric
Rating:
First let's get the Modulus `N`.
```console
$ openssl rsa -in public.pem -text -inform PEM -pubin -text
Public-Key: (256 bit)
Modulus:
00:be:5f:67:0c:7c:df:cc:0b:d3:41:12:d3:bd:71:
22:9f:d3:e4:46:e5:31:bf:35:16:03:6c:12:58:33:
6f:6c:51
Exponent: 65537 (0x10001)
writing RSA key
-----BEGIN PUBLIC KEY-----
MDwwDQYJKoZIhvcNAQEBBQADKwAwKAIhAL5fZwx838wL00ES071xIp/T5EblMb81
FgNsElgzb2xRAgMBAAE=
-----END PUBLIC KEY-----
```
Now convert that into integer.
```console
$ python3 -c "print(int(('00:be:5f:67:0c:7c:df:cc:0b:d3:41:12:d3:bd:71:22:9f:d3:e4:46:e5:31:bf:35:16:03:6c:12:58:33:6f:6c:51'.replace(':', '')), 16))"
86108002918518428671680621078381724386896258624262971787023054651438740237393
```
Since N is small, so it can be factorised easily. By using [FactorDB](http://factordb.com) we can get p and q.
`p = 286748798713412687878508722355577911069`
`q = 300290718931931563784555212798489747397`
Now we have `N`, `p`, `q` and `e` we can calculate the private exponent `d` easily.
`d = (e**-1) mod (p-1)*(q-1)`
```python
e = 65537
p = 286748798713412687878508722355577911069
q = 300290718931931563784555212798489747397
d = pow(e, -1, (p-1)*(q-1))
print(d)
```
which outputs
``` 52563235496868154743721179285926106867856121268586368115409795819089744895137```.
Now we have to make a PEM file out of this. After searching for a long time I can't get a way through it (I tried `RSA.construct()` method several times but it gave me an error). So I decided to make this into a DER file and then convert it to a PEM and then decrypt the flag.
One way to do this is to generate a DER encoded key using OpenSSL's `asn1parse` command's `-genconf` option. But first we need to make an input file and calculate some more values, specifically, `d mod (p-1)` called `e1`, `d mod (q-1)` know as `e2` and `(q**-1) mod p` called as `coeff` (short for coefficient).
```python
e = 65537
p = 286748798713412687878508722355577911069
q = 300290718931931563784555212798489747397
d = pow(e, -1, (p-1)*(q-1))
print('e1 =', d % (p-1))
print('e2 =', d % (q-1))
print('coeff =', pow(q, -1, p))
```
```
e1 = 158375364557874163650127107177698912781
e2 = 13741426462561038188960145920360571165
coeff = 43340310015875206124799642386915239847
```
Now the input file will be :
```console
asn1=SEQUENCE:rsa_key
[rsa_key]
version=INTEGER:0
modulus=INTEGER:86108002918518428671680621078381724386896258624262971787023054651438740237393
pubExp=INTEGER:65537
privExp=INTEGER:52563235496868154743721179285926106867856121268586368115409795819089744895137
p=INTEGER:286748798713412687878508722355577911069
q=INTEGER:300290718931931563784555212798489747397
e1=INTEGER:158375364557874163650127107177698912781
e2=INTEGER:13741426462561038188960145920360571165
coeff=INTEGER:43340310015875206124799642386915239847
```
I named it as `derfile.txt`. Now to make a DER file, run :
```console
$ openssl asn1parse -genconf derfile.txt -out private.der
```
Let's convert DER to PEM :
```console
$ openssl rsa -inform DER -outform PEM -in private.der -out private.pem
writing RSA key
```
Now finally decrypt the flag :
```console
$ openssl rsautl -decrypt -inkey private.pem -in encrypted.txt
AFFCTF{PermRecord}
```
Voila ! We get the flag !
Flag : `AFFCTF{PermRecord}`
