Integral Domain: Solving a^2=1 with at Most 2 Solutions

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Homework Statement



Let R be an integral domain with identity element 1. Show that there at most two elements "a" in R such that a^2=1


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The Attempt at a Solution



Being an integral domain implies that if ab=0, then a=0 or b=0. a^2=1 implies a=1. Then (a)(a) = (1)(1). This seems like a simple problem but I'm not sure how to properly state that there are "at most 2 elements."
 
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Saying that a2=1, is thesame as saying that a2-1=0. Maybe you can factorize the left hand side...
 
Thank you. Got it
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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