, 2018

First assume $n$ is a prime number. Thus $\varphi(n)=n-1 \mid n^{2}+3=(n-1)(n+1)+4$, implying $n-1 \mid 4$. We deduce that $n=2,3$ or $5$.

From now on, assume $n$ is composite and set $n=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$, where $k$ is the number of distinct prime divisors of $n$. Since $n \geq 3$, $\varphi(n)$ is even. In particular, the condition $\varphi(n) \mid n^{2}+3$ shows that $n$ is odd and so the $p_{i}$'s are all odd. It follows that
$$2^{k} \mid \varphi(n)=\prod_{i=1}^{k} p_{i}^{\alpha_{i}-1}(p_{i}-1).$$
So $n^{2}+3$ is divisible by $2^{k}$ implying that $k \leq 2$. Moreover, if $\alpha_{i} \geq 2$ then $p_{i} \mid \varphi(n)$, so $p_{i} \mid n^{2}+3$, forcing $p_{i}=3$. In this case, $9 \nmid n^{2}+3$ and $9 \nmid \varphi(n)$. This shows that $\alpha_{i}=2$. Our discussion shows that if the composite $n \geq 3$ satisfies the problem then $n=3^{2}=9$ or $n=9p$, or $n=pq$ for some odd primes $p, q$. Of course, $n=9$ is a solution of our problem.

- Suppose $n=9p$. Then $\varphi(n)=6(p-1)$ and $n^{2}+3=81(p-1)(p+1)+84$. Therefore, $3(p-1) \mid 84$, or equivalently $p-1 \mid 28$. It follows that $p=5$ or $p=29$. But in both cases, we have $8 \mid \varphi(n)$, contradiction.

- Suppose $n=pq$, where $p, q$ are odd primes. First, consider the case $q=3$ (the case $p=3$ is of course similar). Then $\varphi(n)=2(p-1)$ so $n^{2}+3=9(p-1)(p+1)+12$. It follows that $2(p-1) \mid 12$, and $p-1 \mid 6$, implying $p=3$ or $p=7$. In this case we obtain a new solution $n=pq=3 \times 7=21$ of our problem.

Now, suppose $p, q>3$. Thus, $3 \nmid n^{2}+3$ so $3 \nmid (p-1)(q-1)$. This shows that $p, q \equiv 2 \pmod{3}$. Since $(p-1)(q-1) \mid p^{2}q^{2}+3=(p^{2}-1)(q^{2}-1)+p^{2}+q^{2}+2$ so $(p-1)(q-1) \mid p^{2}+q^{2}+2$. Set
$$p^{2}+q^{2}+2=k(p-1)(q-1),$$
where $k$ is a positive integer. Set $p=2x+1$, $q=2y+1$, then the above equation reads $x^{2}+y^{2}+x+y+1=kxy$, or equivalently,
$$x^{2}-(ky-1)x+y^{2}+y+1=0.$$
By a Vieta jumping technique, we deduce that $k=5$. Thus, we have
$$5(p-1)(q-1)=p^{2}+q^{2}+2.$$
But then $p \equiv q \equiv 2 \pmod{3}$ and the left hand side is $\equiv 2 \pmod{3}$, while the right hand side is $\equiv 1 \pmod{3}$, contradiction.

Therefore, all numbers satisfy the given condition are $n \in \{1,2,3,5,9,21\}$.