Let $\mathfrak N$ be the structure of the natural numbers on the language of arithmetic $\mathcal L=\{0,1,+,\cdot,<\}$. Let $\mathfrak M$ be any nonstandard model of arithmetic. Show that there is an element $0\ne m\in|\mathfrak M|$ such that for every $n\in \Bbb N_+$, we have that
$$\mathfrak M \vDash \exists x(\underline n \cdot x = y)[s(y|m)]$$
where $s$ is any variable assignment.
I've tried thinking of some theorems of arithmetic which might be relevant. This would be any first-order fact, true for $\mathfrak N$. Then any nonstandard model of arithmetic would also have to satisfy this theorem.
I've seen this basic strategy exploited to show that every number is even or odd (including the infinite elements). Then this can be used to show that, for every equivalence class of elements (i.e. $a\sim b$ iff they are a natural number distance apart) there is a smaller equivalence class, and between any two equivalence classes there is a third.
I tried thinking of how we could prove the theorem that, for any $n$, every number is congruent to 0 or 1 or … or $n-1$ mod $n$, basically. However, I haven't seen a way to make use of this. Certainly if $a = n\cdot y + i$ for some $0\le i\le n-1$ then we would have that $a-i$ is a multiple of $n$. But then this makes the quantity $a-i$ dependent on $n$, whereas the exercise asks for a single $m\in|\mathfrak M|$ which works for every choice of $n$.
Since the formula $\exists x(\underline n \cdot x = y)$ is not true for natural numbers $y|m$, we cannot prove this statement as a theorem of arithmetic. And I'm also not seeing another relevant theorem which would help.
I also know that if a formula $\phi(x)$ is true for infinitely many natural (i.e. there are infinitely many $n\in\Bbb N$ such that $\phi(n)$ is true) then there is some infinite element $m\in |\mathfrak M|$ such that $\phi(m)$ is true. But this seems to again get me what I'm not after (as far as I can see). Namely, if you fix $n$ first, then since there are infinitely many multiples of $n$ then there is an infinite element which is a multiple of $n$. However, we still don't know that there is an element which is a multiple of every $n$.
Update: It just now occurs to me, perhaps I can extend the language to a new constant, have an indexed family of sentences which says of this element that it is a multiple of $n$, and then appeal to compactness. I'm going to go try to work out the details of this idea now.
Update 2: Eh, that won't work. It'll show that some nonstandard model of arithmetic has the stated property. But it won't show that every one will.
Best Answer
Your idea of looking for a suitable theorem of arithmetic is the right approach. We can take the following:
$$\forall n \in \mathbb{N} \ \exists m \in \mathbb{N} \ \forall 0 < k \leq n \quad k | m$$
This is easy to prove, just take $m = \prod_{1 \leq k \leq n} k$. Now, if you have a non-standard model $\mathcal{M}$ of arithmetic, then there is going to be some $N \in \mathcal{M}$ which is larger than any standard natural number. Applying the above theorem to $N$ yields a number $M$ which is a multiple of everything up to $N$, in particular of every standard number.