Can anyone guide me how to prove this question?

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Can anyone guide me how to prove this question??

Question:
We want to describe via a picture a set of subsets of a square which are something like diagonals, but are not quite the same. We'll call them steep diagonals. One of them, labelled e, is illustrated in the square below; the other 6 are parallel to it

x_e_x_x_x_x_x
x_x_x_x_e_x_x
e_x_x_x_x_x_x
x_x_x_e_x_x_x
x_x_x_x_x_x_e
x_x_e_x_x_x_x
x_x_x_x_x_e_x

State and prove a theorem about under what conditions we can expect that the sums on the positive ( or negative ) steep diagonals are constant, when we're dealing with a square full of consecutive integers starting at 0.
 
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You're supposed to say what you've done before you get help.
 
Please guide me to start - a simple clue please , if possible. Thank you.
I really don't know what to start with for this question.
 
Start by writing out the square, drawing some steep diagonals and then thinking about what you see.
 
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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