Let $\mathscr L$ be the first-order language having two symbols of constants $\underline{0},\underline{1}$ and two symbols of binary operations $\underline{+},\underline{\cdot} \,.\,$ We consider the $\mathscr L$-structure $\mathscr N = (\Bbb N, 0, 1, +, \cdot)$.
Is the set of perfect numbers definable (without parameters) in $\mathscr N$ ?
My guess is "no", because the number $d(n)$ of divisors of a natural number $n$ depends on $n$, and it seems difficult to express $d(n)$ as a function of $n$ in a first-order language. But I don't know how to disprove it rigorously.
Yes, the set is definable, because it is a computable set. The definable sets are exactly the arithmetical sets, and every computable set is arithmetical.
Any suitable formula will be quite ugly, because the standard method for constructing the formula uses the $\beta$ function. However, it would not be too hard to write down a formula.
Essentially, the formula can go like this. A number is perfect if and only if, when we form a strictly increasing finite sequence consisting of the positive proper divisors of the number, and then form the sequence of partial sums of the first sequence, the overall sum of the first sequence is equal to the original number. Using the $\beta$ function, we can manipulate and quantify over finite sequences like these in the language specified.