Is the relation on $\Bbb R \times\Bbb Z$ given by $(a,b) \sim (c,d) \iff (\exists t \ne0)(\exists n \in \Bbb Z)(c = at \land b^2 = 3n+d^2)$ symmetric? 
Check if $R$  defined on $\mathbb R \times \mathbb Z$ is symmetric, when
$$(a,b) \sim (c,d) \iff (\exists t \ne0)(\exists n \in \mathbb Z)(c = at \land b^2 = 3n+d^2)$$

I have attempted to solve this but I got stuck very quickly. If $R$ was symmetric, then it would imply that
\begin{align*}
&(\exists t\ne0)(\exists n \in \mathbb Z)(c = at \land b^2 = 3n+d^2) \\
\implies &(\exists t_2 \ne 0)(\exists m \in \mathbb Z)(a = ct_2 \land d^2 = b^2 + 3m)
\end{align*}
Is enough to set $t_2 = \frac 1 t$ and $3n = -3m$?
 A: Ultimately, yes, that's enough. Let's rephrase this in a bit clearer of a fashion: assume the following is true:
$$(a,b) \sim (c,d), \text{ i.e. } (\exists t \in \Bbb R \setminus \{0\})(\exists n \in \mathbb Z)(c = at \land b^2 = 3n+d^2)$$
We want to show, on the premise of the above, the following is true:
$$(c,d) \sim (a,b), \text{ i.e. } (\exists t' \in \Bbb R \setminus \{0\})(\exists n' \in \mathbb Z)(a = ct' \land d^2 = 3n'+b^2)$$
To show the existence of $t',n'$, we can simply define them and observe that it works. Namely,
$$c = at \iff \frac 1 a = \frac t c \iff a = \frac 1 t c$$
which suggests the choice of $t' = 1/t$. Similarly,
$$b^2 = 3n + d^2 \iff \frac{b^2-d^2}{3} = n \iff \frac{d^2-b^2}{3} = -n$$
which suggests the choice of $n' = -n$. Of course, since $t \ne 0$, then $1/t \ne 0$; likewise, since $n \in \Bbb Z$, then $-n \in \Bbb Z$. It is also obvious by construction that these definitions of $t',n'$ otherwise satisfy the conditions, validating them.
Thus your relation is indeed symmetric.

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.
