Can someone check this for me?

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Find the prime factorization of 6 in Q[sqrt(-1)].

Ans : Since 6 = 2*3
so 6 = (1+sqrt(-1)) (1-sqrt(-1)) *3 Q.E.D.

Do I need to add anything to it? Am I done with this question? Please kindly advise me. Thank you very much.
 
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How do you know those are prime?
 
because 2 and 3 are prime.
 
I mean 1+sqrt(-1), 1-sqrt(-1), and 3 in the field Q(sqrt(-1)).
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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