Al(bc), then either alb or alc

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Homework Statement


al(bc), then either alb or alc
I'm trying to explain why this is true.


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The Attempt at a Solution


a divides bc.
a divides b or a divides c is true because b is just a multiple of c and a will still be a factor.
 
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This statement is true only if a is prime. Use the fact that if gcd (a,b) = c then there is some integers x,y such that ax+by=c
 
It's not true. Counterexample: 4 | 12, but 4 doesn't divide 6 and 4 doesn't divide 2.
 

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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