How Does Modular Arithmetic Relate to Odd Prime Numbers?

Daveyboy
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b>1, p an odd prime

b2 = -1 mod p

if p|b2 + 1
show p = 1 mod 4

I know that the order of b is 4 (mod p)
 
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That's a strange approach to use...
I would do it this way: Since p is odd, and p|a^2 + 1, a must be an even number. a^2 thus = 0 (mod 4), and consequently a^2 + 1, and p are both congruent to 1 (mod 4)
 
haha, oh man it's so obvious, I can't believe I missed it.
 

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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