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I was wondering if someone could give an opinion on some work as I struggle to know when I'm on the right track or not with groups.

So for: $ (\mathbb R,\circ)$ where $ x \circ y := (x+y)^2$

Is there a neutral element?

At first I wanted to solve $ (x \circ e )^2 $ and show there was no possible solution but I wasn't sure if that was a correct way to do it or if I would even know it is not a possible solution for all $x$.

However, I was wondering if now would it be sufficient to use $ x = -1 $ as a counterexample so that it can be seen:

$[(-1)+e]^2=-1$

Which clearly cannot be true as no value within $\mathbb R$ raised to the power of $2$ can give a minus value.

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    You need to find $e$ which satisfies, that for any $x \in \mathbb{R}: x \circ e = x = e \circ x$, so basically you need to find out, if there is any $e$ such that $(x + e)^2 = x = (e + x)^2$ edit: I wanted to tell you that in the 5th line, you should replace that circle for + – 2017-02-12
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    Yes, that would be sufficient. – 2017-02-12

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