I've got an excercise solved by my teacher, it says I've got to prove a relation $R$ of elements in $\mathbb{R}^2$ is a congruence. In the solved exercise he just proved Reflexivity, transitivity and symetry of $R$. Is this enough? As far as I know that proves only it is a relation of equivalence. Is he missing something?
EDITED from comment: It says "Prove $R$ is a congruence with respect to the sum of pairs" btw $(a,b)R(c,d) \Leftrightarrow 3(a−c)+2(b−d)=0$.
An equivalence relation is just a relation which is reflexive, transitive, and symmetric.
A congruence relation, however, is a bit more: it's a relation which respects some structure. Note that this means the phrase "congruence relation" by itself is vague: I have to tell you what structure I want it to respect.
Here's the standard picture. I have some structure $\mathcal{X}$, which in this case means a set $X$ equipped with a specific "operation of interest" $f$ on $X$. A congruence on $\mathcal{X}$ is then an equivalence relation $E$ which respects this operation, in the sense that $$x_1Ey_1, x_2Ey_2, . . ., x_nEy_n\implies f(x_1, x_2, . . . , x_n)Ef(y_1, y_2, . . . , y_n).$$ For example, let $X=\mathbb{N}$, and consider the operation $+$. Then the relation $E$ given by $xEy\iff [{x\over 2}]=[{y\over 2}]$ (where $[\cdot]$ is the floor operation) is an equivalence relation, but not a congruence: since $1E0$ but $1+1$ is not $E$-related to $0+0$.
(And we can similarly define the notion of congruence for sets equipped with multiple operations, or operations and relations, or various further generalizations.)
So in your context, you have a set $X$ - the set of all pairs of reals - and an operation on $X$ - described in your comment to Asaf. You need to show that the equivalence relation given respects that operation - that is, $$\mbox{$(a, b)R(a', b')$ and $(c, d)R(c', d')$ implies $(a, b)+(c, d)R(a', b')+(c', d')$}.$$ If your teacher indeed only proved reflexivity, symmetry, and transitivity, then they gave an incomplete solution; however, showing that it's a congruence isn't too hard.