If $f$ is non-constant, it is clear that the leading coefficient is positive, thus there exists prime $m$ such that $f(m)>1$. Let $r$ be a prime factor of $f(m)$. Then $ \{ f(m) + nr, f(nr) + m \} $ is a cute subset for any $n\ge 2$. When $n=2$, since $f(m)+2r$ is composite, thus $f(2r)+m$ is prime, say $s$. Note there must exist natural $k$ such that $f((2+ks)r)>f(2r)$, and $f(2r)\equiv f((2+ks)r)\pmod s$. Then $ \{ f(m) + (2+ks)r, f((2+ks)r) + m \} $ is cute, contradiction since both are composite. If $f$ is constant, $f(x)=0$ is a solution, clearly works. If $f(x)=\pm 1$, consider $ \{ p,q \}= \{ 5,9 \}$. Otherwise consider some prime $c$ dividing $f(x)$, then choose $ \{ p,q \}=\{c,2c\}$. Thus our only solution is $f(x)=0$.

MK4J wrote: Call a two-element subset of $\mathbb{N}$ cute if it contains exactly one prime number and one composite number. Determine all polynomials $f \in \mathbb{Z}[x]$ such that for every cute subset $ \{ p,q \}$, the subset $ \{ f(p) + q, f(q) + p \} $ is cute as well. Proposed by Valentio Iverson (Indonesia) Solution. We claim that the only solution is $f(x) = 0$. If $f$ is constant, then suppose $f(x) = c$. Now, $c = 0$ is clearly a solution. If $|c| = 1$, then we can consider the set $\{5,9\}$, which is not cute. If $|c|\neq 1$, then suppose $p$ is a prime dividing $c$. Then considering the set $\{p,2p\}$ gives that $\{p+c,2p+c\}$ is cute. But $c\neq 0$, so $p+c,2p+c$ are both composite, so this is a contradiction. Now suppose $f$ is not constant. Let $p$ be a prime dividing some $f(a)$. Now the sequence $a+kp$ cannot contain all primes, so we can assume that $a>0$ and $a$ is composite (as $p\mid f(a+kp)$). Consider the set $\{p,a\}$. Therefore $\{f(p) + a, f(a) + p\}$ is cute. Suppose $f(a) + p$ is prime, then $f(a) = 0$. But if $f(a+kp) = 0$ for infinitely many $k$ such that $k\in \mathbb{N}$, then $f$ is constant which is false. Therefore we can assume $f(a+kp)\neq 0$ for all sufficiently large $k$. Therefore $f(p) + a+kp$ is prime for all sufficiently large $k$ but this is false.

Does this work- choose some prime $p$ such that $p \mid a_0$, where $a_0$ is the coefficient of the term $x^0$ of the polynomial. Now choose $(p,kp)$, $k>1$. Clearly $p \mid $ both $f(p)+kp$ and $f(kp)+p$, and as $k \rightarrow $ infinity and the polynomial is non-constant, both of our numbers must be composite. If the polynomial is a constant one, clearly $f \equiv 0$ is the only polynomial that works $=>$ we’re done.

Take a prime number $p$ such that $p|f(-f(0))$. Observe that there are infinitely many composite number $q$ such that: $-q\equiv f(0)\equiv f(p)\pmod{p}$ holds.[There exists infinitely many composite in a fixed residue class $\pmod{p}$] Then, $f(q)\equiv f(-f(0))\equiv 0\equiv -p\pmod{p}$. So we have for prime $p$ infinitely many composite $q$ such that: $p|p+f(q)$ and $p|q+f(p)$.So exactly one of them will be $p$ to maintain the property of cuteness. Now $f(p)+q=p$ can hold for at most one value of $q$.So $f(q)+p=p$ will hold for infinitely many $q$.So $f$ has infinitely many zeroes.So $f\equiv 0$.It is indeed a solution.

here's my short solution take $q$ not a prime and $p \in \mathbb{P} : p|f(q)$ suppose $|f(q)| \neq 1$ if $f(q) \neq 0 $ then $p+f(q)$ is not a prime so $f(p)+q \in \mathbb{P}$ since $p|f(q) \implies p|f(q+pq(f(p)+q))$ but $f(p)+q+pq(f(p)+q)=(f(p)+q)(pq+1)$ a contradiction so $f(q)=0$ or $|f(q)|=1$ if $|f(x)| \equiv 1$ then take $\{5,15\}$ is cute but neither $\{4,14\}$ nor $\{6,16\}$ is cute so $f(x) \equiv 0$ and we win

An analytical approach is also possible. Let's see $f$ cannot be a non constant polynomial. Take any prime $p$ such that $m:=|f(0)+p|>1$. Consider the set $$A:= \{q : 1\le q\le N, q \text{ and } f(p)+q \text{ are both composite} \}$$By some straightforward arguments, when $N$ is large enough, the density of $A$ in $[1..N]$ can be made as close to $1$ as we want. By the given requirement it follows $f(q)+p$ is prime for any $q\in A$. But certainly $f(q)+p$ is not prime for $q=km$, where $k\in\mathbb{N}$ is large enough (since $f$ is non constant). It remains to see the density of $\{km : k\in\mathbb{N}\}$ in $[1..N]$ is something close to $1/m$ for large enough $N$.

Similar in idea to some of the solutions above - posting for storage. Assume that $f$ is non-constant otherwise the only solution is $f(x)=0$ which clearly works. Notice that if $2$ divides the constant coefficient of $f$, taking $p=2$, $q=2k$ for $k>1$ gives us that $\{f(2)+2k,f(2k)+2\}$ must be cute, yet this is not possible as we can see that both values are divisible by $2$ and will both get arbitrarily large/small in which case both will be composite. On the other hand if $2$ does not divide the constant coefficient of $f$, then taking $q=2f(p)$ gives us that $\{3f(p),p+2f(p))\}$ has to be cute, as $f$ is non-constant $3f(p)$ will eventually be composite for sufficiently large primes $p$ meaning that $p+f(2f(p))$ has to be a prime, yet $f(2f(p))$ and $p$ are both odd meaning that $p+f(2f(p)) = 2$ for infinitely many primes $p$ which is not possible. We ultimately get that the only solution is $f(x) = 0$. $\blacksquare$