Is there an easier way to prove the divisibility of integers?

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Well, consider the sets of integers (1,...,n) (n+1,...,2n) and so on (and the negatives likewise). Now, these partition the integers. Take n consecutive integers anywhere in the integers and for none of them to divide n exactly you would have to fit them inside one of these sets. That's impossible of course. For two to divide n, you would have to overlap mn and (m+1)n for some integer m. That's also impossible without at least n+1 integers. QED.
 
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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