TEMP="$(mktemp).go"
cat > "$TEMP" <<EOF
package main
import ("crypto/aes"; "crypto/cipher"; "encoding/hex"; "os")
func main() {
  var key, nonce, ciphertext, plaintext []byte; var block cipher.Block; var aesgcm cipher.AEAD; var err error
  if len(os.Args) < 4 { panic("usage: go run salamander.go   ") }
  if key, err = hex.DecodeString(os.Args[1]); err != nil { panic(err.Error()) }
  if nonce, err = hex.DecodeString(os.Args[2]); err != nil { panic(err.Error()) }
  if ciphertext, err = os.ReadFile(os.Args[3]); err != nil { panic(err.Error()) }
  if block, err = aes.NewCipher(key); err != nil { panic(err.Error()) }
  if aesgcm, err = cipher.NewGCM(block); err != nil { panic(err.Error()) }
  if plaintext, err = aesgcm.Open(nil, nonce, ciphertext, nil); err != nil { panic(err.Error()) }
  if _, err = os.Stdout.Write(plaintext); err != nil { panic(err.Error()) }
}
EOF
go run "$TEMP" 5c3cb198432b0903e58de9c9647bd241 4a4f5247454c424f52474553 polyglot.enc > first.jpg
go run "$TEMP" df923ae8976230008a081d23205d7a4f 4a4f5247454c424f52474553 polyglot.enc > second.bmp

This attack was shown by Yevgeniy Dodis, Paul Grubbs, Thomas Ristenpart, and Joanne Woodage in Fast Message Franking: From Invisible Salamanders to Encryptment.

First, we will describe a general strategy to create a ciphertext that yields the same MAC with two different keys. Then we will show how to construct a ciphertext that yields meaningful results when decrypted with those two keys.

Consider arbitrary keys \(k_1, k_2\), nonce \(n\), and ciphertext \(c\) (additional associated data can be accounted for in a straightforward manner). \(k_1, k_2\) are associated with authentication keys \(h_1, h_2\) and blinds \(s_1, s_2\), respectively.

Given a ciphertext of three blocks, we will attempt to add a fourth ciphertext block \(c_3\), set the MACs equal to each other, and solve for \(c_3\). Remember that in \(\mathbb{F}_{2^{128}}\), addition is the same as subtraction.

For the resulting ciphertext of four blocks, the MACs for each key are computed as \[ mac_{1} = s_1 + \vert c\vert{}h_1 + c_3h_1^2 + c_2h_1^3 + c_1h_1^4 + c_0h_1^5 \] \[ mac_{2} = s_2 + \vert c\vert{}h_2 + c_3h_2^2 + c_2h_2^3 + c_1h_2^4 + c_0h_2^5 \] where \(s\) is a constant depending on the AES-GCM key and the nonce, and \(h\) is the authentication key depending only on the AES-GCM key. We can now set the MACs equal to each other and solve for \(c_3\). \[ s_1 + \vert c\vert{}h_1 + c_3h_1^2 + c_2h_1^3 + c_1h_1^4 + c_0h_1^5 = s_2 + \vert c\vert{}h_2 + c_3h_2^2 + c_2h_2^3 + c_1h_2^4 + c_0h_2^5 \] \[ c_3(h_1^2 + h_2^2) = (s_1 + s_2) + \vert c \vert (h_1 + h_2) + c_2(h_1^3+h_2^3) + c_1(h_1^4 + h_2^4) + c_0(h_1^5 + h_2^5) \] \[ c_3 = \frac{(s_1 + s_2) + \vert c \vert (h_1 + h_2) + c_2(h_1^3+h_2^3) + c_1(h_1^4 + h_2^4) + c_0(h_1^5 + h_2^5)}{h_1^2 + h_2^2} \]

Note that the choice to place the extra block in the final position was arbitrary. For the attack below we will instead need to change the penultimate block rather than adding a block; the computation is similar.

For the next phase, we construct a ciphertext that decrypts to a valid JPEG under one key and a valid BMP under another. The basic strategy is to place the JPEG bytes and BMP bytes at different locations, carefully arranging it so each parser will ignore the other data for the file. JPEG files can include comments, in which we will include the BMP data. The BMP parser will stop reading as soon as the indicated length of the BMP has been read, after which we will include the JPEG data. In each decrypted file, the data for the other image will be scrambled as we are using a different key, but it will not matter as the garbage data will be in a location that is ignored by the image parser.

All JPEG files start with the magic bytes ffd8 and end with ffd9. We will place a JPEG comment immediately after the initial magic bytes, which is indicated by fffe and is followed by a 2-byte big-endian encoding of the comment length. All BMP files start with the magic bytes 424d followed by a 4-byte little-endian encoding of the file length.

Because the JPEG comment has a maximum length, our BMP file will need to be less than ffff=65536 bytes. Thus, we desire the JPEG file to start with ffd8 fffe ffff and the BMP file to start with 424d wxyz 0000, where wxyz is the actual length of our BMP file.

To make it easier, we will only require the first five bytes to match; we can modify ciphertext afterwards to satisfy the final byte of the BMP header. The final byte of the JPEG header will be arbitrary, but this is the least significant part of the comment length, so we will restrict our BMP file length to less than ff00=65280 bytes.

Since these must be in the same location at the start of the file, we will need to brute-force search for two keys \(k_1, k_2\) and a nonce that encrypt to the same ciphertext. The easiest way to do this is via a birthday attack: fix an arbitrary nonce, then generate random keys for both the JPEG header and the BMP header. Encrypt each and store them in a lookup table. Repeat until two keys are found that encrypt their respective headers to the same ciphertext bytes.

To the ciphertext header, add the encryption of the BMP file (without the header) under \(k_2\), then pad with arbitrary data to reach the end of the JPEG comment (this will be ignored by the BMP parser, which has already finished reading the file). After the JPEG comment is over, add the encryption of the JPEG file (without the header and final magic bytes) under \(k_1\).

Finally, add two more bytes of ciphertext that will make the final two bytes of the JPEG file into the appropriate final magic bytes ffd9.

These ciphertexts do not have the same MAC yet. If we tried to use the strategy outlined at the beginning where we add an extra block at the end, the JPEG file would no longer end in ffd9 and thus would be invalid. Instead, we need to modify it to change the penultimate block.

However, we don't want any data from the penultimate block to show up in our JPEG image. Thus, after the JPEG file data ends, we start another comment that will extend until the penultimate block, hiding it from the parser. Care must be taken to ensure the penultimate block is really on a block boundary (the comment can be padded to increase the length if necessary). After this second comment will appear the final magic bytes ffd9 as desired.

from aesgcmanalysis import att_merge_jpg_bmp

with open('first.jpg', 'rb') as h:
    jpg = h.read()
with open('second.bmp', 'rb') as h:
    bmp = h.read()
c, mac = att_merge_jpg_bmp(jpg, bmp, aad=b"")