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 5 of 7: Substitute the affine form back into the equation

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In plain words

So far only two special substitutions were used; the full equation must still be checked and will constrain k and c further.

f(2a)+2f(b)=2k(a+b)+3c,\quad f(f(a+b))=k^2(a+b)+kc+c\ \Longrightarrow\ 2k=k^2,\ \ 2c=kc

Detailed analysis

Write f(x) = kx + c. Substituting into the original equation, the left side f(2a) + 2f(b) equals 2k(a+b) + 3c, while the right side f(f(a+b)) equals k squared times (a+b) plus kc plus c. Since this identity must hold for every choice of integers a and b, the coefficients of (a+b) must match and the constant terms must match, giving the two conditions 2k equals k squared, and 3c equals kc plus c.