In my attempts, I'm trying to write all the details that I understand, since mathematical logic seems very mysterious to me (like sometimes you don't even have addition and have to define one???)
Let $M_1 = (\mathbb{Z},+)$ and $M_2 = (\mathbb{Z},+,\cdot)$.
1) Show that $f: x \rightarrow -x$ is an automorphism of $M_1$. Deduce that there is no formula $\phi(x)$ such that $M_1 \vDash\phi(a)$ if and only if $a$ is positive.
Attempt: The first part is a routine check of the definitions of automorphism. But I got stuck badly for the second part. Suppose there is a formula $\phi(x)$ such that $M_1 \vDash \phi[a]$. Then we have $M_1 \vDash \phi[f(a)]$, where $f$ is the map in the first part (there is a theorem that asserts this. But this just means, $M_1 \vDash \phi[-a]$. Then I just don't see anything I can get from this direction...
2) Show that there is a formula $\psi(x)$ such that $M_2 \vDash \psi[a]$ if and only if $a$ is positive. (Hint: Use the Lagrange's four square theorem).
Attempt: First we enrich our language by defining squaring a number. Let $\tau[y]=y \cdot y$ and from now on we define the formula $\tau[y]$ by $y^2$. Next we let $\psi[x]=\exists t\exists u\exists v\exists w(x=t^2+u^2+v^2+w^2)$. Then if $a$ is a positive number $\mathbb{Z}$, we have $M_2 \vDash \psi(a)$ by the Lagrange's Four Square Theorem. Conversely, suppose there is an $a$ such that $M_2 \vDash \psi(a)$, then since $a$ is equal to a sum of four squares, it's true that $a \geqslant0$. Is this proof correct, and if so, how do I exclude the case $a=0$?
Thanks!