MathLabs

Problem 2

Let S={2,3,4,…}S=\{2,3,4,\ldots\} denote the set of integers that are greater than or equal to 22. Does there exist a function f:S→Sf:S\to S such that f(a)f(b)=f(a2b2)f(a)f(b)=f(a^2b^2) for all a,b∈Sa,b\in S with a≠ba\ne b?
Step 1 of 4: Compare two factorizations
f(a4b4c4)=f(a2)f(b)f(c)=f(b2)f(a)f(c)f(a^4b^4c^4)=f(a^2)f(b)f(c)=f(b^2)f(a)f(c)
Detailed analysis

For arbitrary a and b choose c larger than both. Applying the given equation twice to the same fourth-power product is legal because the unequal arguments condition is satisfied at each application. Cancelling the positive value f(c) gives the displayed relation.