Could you help me with that, I don't know how to prove if the relation is an equivalence and the class of 5?
On the set of integers, the relationship is defined by $aRb \iff a^2-b^2=a-b$. Find out if it is an equivalence relation and where appropriate, find the equivalence class of 5.
First, the relation is clearly reflexive since $a^2 - a^2 = 0 = a - a$ for any integer $a$.
Second, the equation $a^2 - b^2 = a - b$ is symmetric since switching $a$ and $b$ introduces negative signs on both sides. More precisely, if $a^2 - b^2 = a - b$ then $b^2 - a^2 = -(a^2 - b^2) = -(a - b)= b-a$.
Is it transitive? Suppose that $a^2 - b^2 = a - b$ and $b^2 - c^2 = b - c$. Then $a^2 - c^2 = a^2 - b^2 + b^2 - c^2 = (a^2 - b^2) + (b^2 - c^2) = (a-b) + (b - c) = a - c$, so indeed it is transitive.
To find the equivalence class of $5$ we need to find all integers $a$ such that $a^2 - 5^2 = a - 5$ which is equivalent to $0=a^2 - a - 20=(a + 4)(a-5)$, so we see that $a = -4$ or $a = 5$. Hence the equivalence class of $5$ consists of $\{ -4, 5\}$.
$a R b$ iff $f(a)=f(b)$, where $f(x)=x^2-x$. Since equality is an equivalence relation, so is $R$.
It is necessary to solve the equation: $$a^2-b^2=a-b$$
The decision will then be:
$$a=\frac{-k+n+1}{2}$$
$$b=\frac{k-n+1}{2}$$
$k,n$ - integers different parity. Any sign.
