A nice problem

Since the interval is symmetric about 0, you're just asking for how many numbers in [-2007, 2007] are quadratic residues mod 2^2007.

sum(n=-2007,2007,issquare(Mod(n,2^2007))) = 670 (15ms in Pari)

Now if you actually want to solve it by hand, you'll need to look over quadratic reciprocity more carefully.

 

Wikipedia, under "quadratic residue", says that a number is a residue mod 2^m (m any natural number) if and only if it is of the form (4^k)*(8n+1).
I counted 336 integers in [1,2007] that are of that form. I thought an equal number might be in [-2007,-1], but that would give me 672, not 670.

 

Is it at this point that quadratic reciprocity should be applied? But how if it applies to odd primes?
Something else (in my previous post): m any natural number > 2.

 

Using the form -(4^k)*(8n-1), I can count 333 residues in [-2007,-1]. With 0 also a residue, I get 670.