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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.

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    Of course you can. It's a recursive set. – 2017-01-23
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    @Jonathan : ok, it is recursive because for every $d < n$, you check whether it divides $n$, and if so you add it to the previous divisor of $n$ you checked (starting with $d=0$). Then you check whether the final sum is $n$. But my question is then: can you give my an explicit formula that shows it is definable? – 2017-01-23
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    (By the way, do we already know if it is representable by a $\Delta_0^0$ formula, or $\Sigma_1^0$ or $\Pi_1^0$?) – 2017-01-23
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    As @Jonathan says, this is recursive, so we should be able to do this ,... but you'll have to use Godel's $\beta$-function to do this, so it won't be pretty if you actually try! – 2017-01-23
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    @Bram28 : I see... But recursivity of the set $P$ of perfect numbers ensures that there is a $\Sigma_1^0$ formula representing $P$, or is it even a $\Delta_0^0$ formula? – 2017-01-23
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    @Watson ... Not familiar with your sigma or delta notation ... Does that stand for first-order or second-order logic? Anyway, using the $\beta$-function there is a first-order logic formula for this. – 2017-01-23
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    @Bram28 : this refers to [arithmetical hierarchy](https://en.wikipedia.org/wiki/Arithmetical_hierarchy#The_arithmetical_hierarchy_of_formulas). – 2017-01-23
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    @Watson Thanks ... I'll need to take a close look at that to say what class it belongs to ... – 2017-01-23
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    @Watson: if a set is recursive, then it is $\Delta^0_1$ definable, which is to say both $\Sigma^0_1$ and $\Pi^0_1$ definable (by different formulas, of course). There will not be a $\Delta^0_0$ formula - that would mean a quantifier-free formula, and the sets definable by a quantifier-free formula in the given language are quite simple. – 2017-01-24

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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.