MathLabs

Problem 1

Let Z\mathbb{Z} be the set of integers. Determine all functions f ⁣:Z→Zf\colon\mathbb{Z}\to\mathbb{Z} such that, for all integers aa and bb, f(2a)+2f(b)=f(f(a+b)).f(2a) + 2f(b) = f(f(a + b)).
Step 3 of 7: Derive additivity
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In plain words

Substituting the doubling identity into another pair of instances of P cancels the outer f and leaves a plain additive relation.

f(a)+f(b)−f(0)=f(a+b)for all a,b∈Zf(a)+f(b)-f(0)=f(a+b)\quad\text{for all }a,b\in\mathbb{Z}
Detailed analysis

Comparing P(a, b) and P(0, a+b), both equal f(f(a+b)): the first gives f(2a) + 2f(b), and the second gives f(0) + 2f(a+b). Replacing f(2a) using the identity from the previous step and simplifying shows that f(a) + f(b) minus f(0) equals f(a+b), for all integers a and b.