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 6 of 7: Solve for k and c

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In plain words

The condition on k factors, splitting the problem into exactly two cases, and each case then determines whether c is forced or free.

k(k-2)=0\ \Longrightarrow\ (k=0,\ c=0)\ \text{ or }\ (k=2,\ c\in\mathbb{Z}\text{ free})

Detailed analysis

From 2k equals k squared we get k times (k minus 2) equals zero, so k equals 0 or k equals 2. If k equals 0, the condition 2c equals kc becomes 2c equals 0, forcing c equals 0. If k equals 2, the condition becomes 2c equals 2c, which holds for every integer c, so c is unrestricted in this case.