Let $\Sigma$ be an alphabet, $u,v \in \Sigma^*, u \neq v$.
$G_{u,v} = \{ \langle M \rangle \mid M$ is turing machine, $u \in L(M) \Leftrightarrow v \in L(M)\}$.
I understand that if it is not recognizable then its also clearly not decidable. I guess by assuming it to be recognizable one can construct some impossible machine...
On
Using Rice's theorem you may claim that it is not recognizable (as any machine accepting the empty set is in the language and this is a 'semantic' language, determined only by the languages of the Turing machines involved). For the other direction, to show that its complement is not recognizable, you can show a reduction from $A_{TM}$ to this language, that is, $$f(<M,w>)=N$$ such that on input $x\neq u$, $N$ accepts, and on input $u$, $N$ simulates the run of $M$ on $w$ and answers according to that run.
One could also use Rogers's fixed point theorem (https://en.wikipedia.org/wiki/Kleene%27s_recursion_theorem#Rogers's_fixed-point_theorem) to prove this, although the earlier answer is more general.
Assuming for a contradiction that $G_{u,v}$ were recognizable, say by a Turing machine $M_c$, then by Church's thesis (https://en.wikipedia.org/wiki/Church%E2%80%93Turing_thesis#Informal_usage_in_proofs), there is a recursive function $F$ such that for all $x$, $L\left(M_{F(x)}\right) = \{u\}$ if $M_c$ accepts $x$ and $L\left(M_{F(x)}\right) = \emptyset$ if $M_c$ does not accept $x$. By Rogers's fixed point theorem, there is some $x_0$ (in fact, infinitely many such $x_0$) such that $L\left(M_{F(x_0)}\right) = L\left(M_{x_0}\right)$.
Now we get a contradiction: if $M_c$ accepts $x_0$, then $L\left(M_{x_0}\right) = \{u\}$, but this implies $u \in L\left(M_{x_0}\right)$ and $v \notin L\left(M_{x_0}\right)$, contradicting the assumption that $x_0 \in L\left(M_c\right)$. If $M_c$ does not accept $x_0$, then $L\left(M_{x_0}\right) = \emptyset$, that is, $u \notin L\left(M_{x_0}\right)$ and $v \notin L\left(M_{x_0}\right)$, implying $x_0 \in L\left(M_c\right)$, which is again a contradiction.