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 4 of 7: Solve the additive equation over the integers

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In plain words

A function that is additive on the integers is completely determined by its value at 1.

g(x):=f(x)-f(0)\ \Longrightarrow\ g(n)=n\,g(1)\ \Longrightarrow\ f(x)=kx+c,\ k=g(1),\ c=f(0)

Detailed analysis

Define g(x) = f(x) - f(0). The previous step says g(a+b) = g(a) + g(b) for all integers a and b. Setting a = b = 0 gives g(0) = 0, and an induction on n using g(n+1) = g(n) + g(1) together with g(-n) = -g(n) shows g(n) = n times g(1) for every integer n. Hence f(x) equals k x plus f(0) for every integer x, where k denotes g(1); that is, f is forced to be an affine function of x.