Number of plaintext–ciphertext pairs needed to rule out false keys with a reasonable likelihood in AES I am new to cryptography and I've got the following question from Understanding Cryptography:
Imagine that aliens — rather than abducting earthlings and performing strange experiments on them — drop a computer on planet Earth that is particularly suited for AES key searches. In fact, it is so powerful that we can search through 128, 192 and 256 key bits in a matter of days. Provide guidelines for the number of plaintext–ciphertext pairs the aliens need so that they can rule out false keys with a reasonable likelihood.
What I tried
I know that the expected number of keys that encrypt a plaintext $x_1$ to $y_1$ is $2^{k-n}$, where $k$ is the key length in bits, and $n$ is the size of a block in bits.
I also know that given $\{(x_1, y_1), \dots, (x_t, y_t)\}$ PT-CT pairs, the expected number of false keys which encrypt all plaintexts to the corresponding ciphertexts is $2^{k-tn}$.
Therefore, for AES-128, AES-192, and AES-256, I would expect that only two PT-CT pairs is sufficient, given that the expected number of false keys would be, respectively, $2^{-128}$, $2^{-64}$, $2^0=1$.
What the answer manual says
The manual says that, for AES-128, only one pair is sufficient. This one I can see why, since the expected number of keys that map $x_1$ to $y_1$ one be $\frac{2^{128}}{2^{128}}=1$.
For AES-192, the manual says that &quot;in order to find the correct key with a probability of 50 percent, we require $2^{63}$ pairs of plaintexts and ciphertexts&quot;. And for AES-256, also with 50% probability, $2^{127}$ PT-CT pairs are required.
I cannot see why such values are required for AES-192 and AES-256.
I cannot see why such values are required for AES-192 and AES-256.

That's because you have a correct understanding of the situation, and the answer manual is wrong.
Your analysis in &quot;what I tried&quot; is correct; there's not a whole lot I can add to it...