Find out a collision: AES(y, x) ⊕ y under key x

[Raman]

This problem is from a cryptography course.
Let ⊕ be the XOR operator.
AES (Advanced Encryption Standard) is an encryption function that takes two arguments (plaintext message, key) each ranging between 0 and 2¹²⁸-1 and returns a ciphertext value again ranging between 0 and 2¹²⁸-1.
For a given key, the encryption from plaintext to ciphertext is easily computable (AES function) and also the decryption from ciphertext to plaintext is easily computable (AES^-1 function).
For a given key, the set of plaintext messages and the set of ciphertext messages are being bijective.
However, with the given plaintext messages and ciphertext messages, computing the key is computationally out of reach.
However, I am neither sure that if for a given plaintext message, there can be two keys that encrypt it to the same ciphertext message (Again that assume that computing to finding out such a collision is computationally out of reach).
nor sure that if for a given ciphertext message, there can be two keys that decrypt it to the same plaintext message (Do these two statements that are being essentially equivalent ?? I do wonder that !!).

Consider these two problems
1. AES(x₁, x₁) ⊕ y₁ = AES(x₂, x₂) ⊕ y₂ with x₁ ≠ x₂ and y₁ ≠ y₂
2. AES(y₁, x₁) ⊕ y₁ = AES(y₂, x₂) ⊕ y₂ with x₁ ≠ x₂ and y₁ ≠ y₂

The first problem is being stupidly easy. Right ?? Isn't it...
It is quite very obvious / trivial / natural that if you fix x₁, x₂, y₁ to arbitrary values, then you could calculate y₂ = AES(x₁, x₁) ⊕ y₁ ⊕ AES(x₂, x₂).

I am looking out for a solution to the second problem... Right ?? Isn't it...
How will you go about for solving the second problem ?? !!
It is quite very subtle that
→ if you fix x₁, y₁, y₂ to arbitrary values, then computing key x₂ from plaintext and ciphertext is being impossible.
→ if you fix x₁, x₂, y₁ to arbitrary values, then we should deal with y₂ variable at two places, one inside of the AES function, one outside of the AES function along with the XOR (⊕) operator up on the right hand side.
→ on the other hand, if the second problem were
AES(y₁, x₁) ⊕ x₁ = AES(y₂, x₂) ⊕ x₂ with x₁ ≠ x₂ and y₁ ≠ y₂
then again that it would have been easier to do solve
are being if you fix x₁, x₂, y₁ to arbitrary values,
then you could calculate y₂ = AES^-1(AES(y₁, x₁) ⊕ x₁ ⊕ x₂, x₂).

Originally Posted by Raman

Mentally stimulating / simulating
Mind stimulating / simulating
Warm up / relax / reflex / refresh / retain / restrain / recreation
exercise / activity / assignment / quiz
sport / game / puzzle
sheet / chart / pape - r / page - r / pape - r / page - r / paupe - r / barbe - r / carpente - r / bank / bankrupt / agent / agency
globe / earth / world

1. Make out 127 by using the digits 1, 2, 7 once with out changing the order.
2. Make out 127 by using the digits 1, 2, 7 twice.
3. Make out 127 by using the digits 1, 2, 7 N times.
4. What is being the largest number that can be made out done by using that 3 digits? done.
5. What is being the largest number that can be made out done by using that 3 threes? done.
6. What is being the largest number that can be made out done by using that 4 ones? done.
7. With out changing the order of the digits, by using the digits 3, 1, 3, 6 make out 8.

Number Twist
My Own Two Cents
Piece of Cake
Bonus
You can use operators / operations for the concatenation, addition, subtraction, multiplication, division, square root, factorial, power (index / exponentiation), parenthesis (brackets), absolute value (modulus).
8. Make out 87 by using the digits 1, 3, 5, 7.
9. Make out [strike]76[/strike] 84 by using the digits 0, 4, 5, 9.
10. Make out 47 by using the digits 0, 1, 2, 3.
Absolute value (modulus) is being essentially useless - → |x| = x or -x. |x - y| = x - y or y - x.