Problem 1

Let

\mathbb{Z}

be the set of integers. Determine all functions

f\colon\mathbb{Z}\to\mathbb{Z}

such that, for all integers

a

and

b

f(2a) + 2f(b) = f(f(a + b)).

Step 7 of 7: Conclude the classification

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In plain words

The two surviving cases are exactly the two families that were checked to work at the very start.

f\equiv 0\qquad\text{or}\qquad f(x)=2x+c\ \text{ for some fixed }c\in\mathbb{Z}

Detailed analysis

The case k equals 0 and c equals 0 gives the constant function f identically zero. The case k equals 2 gives f(x) equals 2x plus c for an arbitrary fixed integer c. Since both families were already verified to satisfy the original functional equation, and the argument above shows no other function can satisfy it, these are exactly all the solutions.