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I encrypt 2 random keys with the same 'one' time pad. You can get both encrypted keys. You don't know anything besides the size.
Is there a way to attack the two time pad in this scenario?
And what if I would encrypt 1000 random keys. is there a way to attack the 'one' time pad?
PS. I have read a lot of posts about plain text attacks and language analysis which both certainly don't apply here.
At least, doing the goof of reusing the OTP makes one vulnerable to disclosure of any of the key, which trivially reveals all the others.
For the rest, the consequences depends heavilly on what the keys are intended for. If the keys are intended for a block cipher that is secure including under related-key attack (as AES almost is), then there is not consequence beyond the aforementioned.
But in other cases the goof can be damaging, even disastrous. Simple example of the worse: the two random keys enciphered under two-times Pad are really random half-keys, intended to be held separately and combined by XOR to form a full key when needed (which is a simple and most common form of secret sharing with 2-out-of-2 parameterization); with the goof, the full key is the XOR of the ciphertexts for the two half-keys, thus is compromised, even though the individual half-keys remain confidential.
There is a spectrum in-between if the keys are for ciphers with poor security, especially under related-key attacks. In particular, if there are many keys enciphered under many-times pad, the goof allows the attacker to effortlessly pinpoint a key pair with much lower Hamming distance than average, and that pair might become the basis of a cryptanalysis (e.g. from ciphertexts for known plaintexts enciphered under both keys).
Let's just consider keys of bitlength 1, it works like that for any length.
How can you attack the OTP? You can't. However, you don't need it. Because $x_1$ and $x_2$ quite obviously reveal a lot about the things you wanted to hide in the first place:
$$x_1 \oplus x_2 = k_1 \oplus k_2$$
So if their "encryptions" are equal, then the keys are equal. Without knowing the "encrypted" keys, there are 4 possibilities ($k_1,k_2 \in \{0,1\}$), and then you reduce it to 2 possibilities (which ones depend on $x_1$ equal to $x_2$ or not).
Yes, this is bad. Over $n$ bit keys, instead of having $2^{2n}$ possibilities for both keys, you only have $2^n$. And you have to store 3 instead of 2 keys.
