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IMO Team Selection Contest

Algebra Difficulty 7.7 National Olympiad, round 2 Prove it Estonia

Does there exist an operation ∗* on the set of all integers such that the following conditions hold simultaneously:
(1) for any integers xx, yy, zz, (x∗y)∗z=x∗(y∗z)(x * y) * z = x * (y * z);
(2) for any integers xx and yy, x∗x∗y=y∗x∗x=yx * x * y = y * x * x = y?

Solution

Define an operation ⊕\oplus on the set of all non-negative integers, which maps two non-negative integers aa and bb to a non-negative integer a⊕ba \oplus b, such that for all i=0,1,…i = 0, 1, \dots, (a⊕b)i=(ai+bi) mod 2(a \oplus b)_i = (a_i + b_i) \bmod 2, where nin_i stands for the binary digit corresponding to 2i2^i in the binary representation of nn. This operation satisfies condition (1) for all non-negative integers because addition modulo 2 satisfies it. The operation also satisfies condition (2) because if x,y∈{0,1}x, y \in \{0,1\}, then (x+x+y) mod 2=y mod 2=(y+x+x) mod 2(x + x + y) \bmod 2 = y \bmod 2 = (y + x + x) \bmod 2.

As the set of non-negative integers as well as the set of all integers are countable, there exists one-to-one correspondence ff between these sets (e.g. mapping a non-negative integer xx to the integer (−1)x⌊x+12⌋(-1)^x \lfloor \frac{x+1}{2} \rfloor). Every integer can therefore be uniquely expressed in the form f(n)f(n), where nn is a non-negative integer. Therefore we can define the operation ∗* by the formula f(x)∗f(y)=f(x⊕y)f(x) * f(y) = f(x \oplus y). Following from the construction, both conditions (1) and (2) still hold.

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