A special kind of multiplicative function $f: \mathbb N \to \mathbb N$ such that $f(p)=p+k$ for all odd prime $p$, where $k>1$ is a fixed odd integer For which odd positive integer $k$, can we find a multiplicative function $f: \mathbb N \to \mathbb N$ satisfying the following conditions :
$f(p)=p+k$ for all large enough odd prime $p$ and the set $\{n \in \mathbb N : n$ is not square free ; $ f(n) $ is perfect square $\}$ is finite.
Does there exist even at least one odd positive integer $k$ for which we can find a function of the above type ? 
 A: Suppose that the Schinzel's hypothesis is true. It suffices to assume that for any pair of coprime integers $a$ and $b$ with $a>0$ and such that $(a,b) \neq (c^2,-d^2)$ for any integers $c$ and $d$ the polynomial $an^2+b$ takes prime values infinitely often. Now we will prove that the function $f$ with stated properties does not exist for any $k$. As $k$ has only finitely many divisors, there exist $d \mid k$ such that the set $$\mathcal F_d=\{n: n=p^2, (f(n),k)=d\}$$ is infinite. Choose any two different $a,b \in \mathcal F_d$. Then $f(ab)=f(a)f(b)=d^2m$ with $(m,k)=1$. Therefore, for large enough prime $p$, we have $f(abp)=x^2$ if and only if $m(p+k)=y^2$. If $s$ is a squarefree part of $m$ and $s\neq 1$, then $p=sn^2-k$ for infinitely many $p$, and so $f(abp)$ is a square for infinitely many $p$, which contradicts the assumption of the question. Therefore, $s$ always equals $1$. Thus, for any $a,b \in \mathcal F_d$ the number $f(ab)$ is always a square, which is also a contradiction.
It is still interesting to prove that no such function exist unconditionally.
