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I guess we need to see a pattern and confirm our pattern by induction.

How about trying to see repeated roots?
0, 0, 1, 2, ...

It is still a useful thought. Maybe working with \(\tan\) would help us achieve something?

Actually it would reserve our sequence to [0,1]...

Oh, that skipped me! Wow!!!

Amazing!

\(2^{1997}\theta\) should be a multiple of \(\pi\)...

I'm really bad at counting. :|

So far, we've calculated 1998 unique values of \(\sin^2\theta\). Is that it?

Oh, nope.

Exactly...

Oh, I got it. Thanks.

I double-counted 0 at \(\theta = 0 \) and \(\pi\). :P

1997 distinct values?

that would be too less when you compose a quadratic 1998 times

also recall the fact that sin is one-to-one in the interval [0, pi/2]

http://gyazo.com/f22b532bddfbe6cd15af74145160b997

Oh my, never mind...

Goddammit.

find the values of \(n\) such that \(\theta \in [0,\pi/2]\)

Yes, alright. So \(2^{1997}\) distinct values.

(?)

Or am I misunderstanding what you're saying?

the solution i have says \(2^{1996}+1\)
so double check..

Oh my god, what is wrong with me...

Obviously, yes.

What are the last three digits of that? I don't know how modular arithmetic works.

Remind me to get some sleep after this one. Thanks.

work \[2^{1996}+1 \pmod{1000}\]

Yes, I have no idea how modular arithmetic works.

eating dinner.. 10 mnts

My ... what a beautiful solution you came up with.

you may use chinese remainder thm or go wid binary exponentiation

i don't see any other neat way to reduce 2^1996

W|A returns 337.