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In any integral domain, only $1$ and $-1$ are their own multiplicative inverses. Note that $x=x^{-1}$ iff $x^2=1$

I'm not sure how to go about proving this.

I know the definition of an integral domain is defined to be a commutative ring with unity having the cancellation property.

This is how I started:

Assume $x=x^{-1}$ in any integral domain, then by multiplying $x$ on both sides we get $x^2=e$ where $e$ is the identity element. However, since we are in an integral domain, then there exists a unity which is the neutral element. Then: $x^2 \times 1=x^2$, so $e=1$ and we have $x^2=1$

Here is another idea I have for the other way around:

Let $x^2=1$ then subtracting $1$, we obtain $x^2-1=0$ which can be factored into $(x+1)(x-1)=0$ Say $x+1=0$ then $x=-1$. If $x-1=0$ then $x=1$ but then I do not know how to get to $x=x^{-1}$

Chilanie
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    The first step is to show that $x=x^{-1}$ if and only if $x^2=1$. Then since we are in an integral domain, the product $(x-1)(x+1)$ is equal to $0$ if and only if $x-1=0$ or $x+1=0$, and we are finished. – André Nicolas Apr 27 '15 at 05:31
  • @AndréNicolas would you mind making your comment an answer? In case you do, would it be OK if I post an answer based on your comment? – evaristegd May 31 '19 at 01:20

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