Question
Determine if
$f(n) = n+k$
(k is a fixed real number)
are completely multiplicative, multiplicative or neither.
Attempted solution
The only background I have at this point are (1) the definitions of completely multiplicative and multiplicative and that (2) completely multiplicative implies multiplicative, so this is a bit back-of-the-envelope.
Here I try a counterexample to the multiplicative assertion with $2$ and $3$ (chosen because they are relatively prime):
$$f (2 \cdot 3) = 6 + k$$ $$f(2) = 2 + k$$ $$f(3) = 3 + k$$
$f(6)$ is only equal to $f(2)f(3)$ if
$$6 + k = (2+k)(3+k)$$ $$6 + k = 6+5k+k^{2}$$ $$0 = 4k+k^{2}$$ $$k(4+k) = 0 $$
This becomes $k_{1} = 0$, $k_{1} = -4$
I draw the conclusion that this function is not multiplicative (and thus not completely multiplicative by definition), because then it would have been true for all k. But there is no reason to suppose that k has these two values regardless of values tested for $m$ and $n$.
I then try the general case:
$$f(n) = n + k$$ $$f(m) = m + k$$ $$f(mn) = mn + k$$
The function is multiplicative if (assuming $m$,$n$ relatively prime)
$$mn + k = (n+k)(m+k)$$ $$mn + k = mn + (n+m)k + k^2$$ $$0 = (n+m-1)k + k^{2}$$ $$k(k+n+m-1)$$
This gives the two solutions of $k_{1} = 0$ and $k_{2} = -(n+m-1) = 1 - n - m$.
However, I am a bit stuck here and do not really know how to wrap this up. What is a reasonable next step and at what point are these types of questions considered solved?
Your $k$ can't depend on $n$ and $m$ (since $k$ is fixed and the equality should be true for all values of $n$ and $m$). I.e. only $k=0$ is possible.