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Let $R$ be the relation defined on $\mathbb{Z}$ where $a\; R\; b$ means that $a + b^2 \equiv 0\pmod{2}$.

How would I go about finding the equivalence class $[-13]$?

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You would find all $a \in \mathbb{Z}$ such that $a+(-13)^2 = 0 \pmod{2}$ since $[-13] = \{a \in \mathbb{Z}: aR -13 \}$.

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    I proved it by showing R to be reflexive, symmetric, and transitive. I just don't have much experience with equivalence classes.2011-01-29
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$a+b^2=0\ (2)$ is the same as "$a$ and $b$ have the same parity", so the set of odds is $-13$'s class.

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    Just to explain what I'm getting at, Krysten asked for 'how' and if she's doing that then she'd probably also need a step or two of 'how' in order to get to your parity statement. That is, -how- do you know that the strange looking sum leads to the parity statement. And that's what Bill's answer helps with. Sorry making so much of not much, I think those missing details are what the OP hoped for.2011-01-30
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HINT $\ \ \rm b^2 \equiv -b\ \ (mod\ 2)\ $ thus $\rm\ a\ R\ b\ \iff\ a\ \equiv\ b\ \ (mod\ 2)\ $ which is an equivalence relation.