-- All square roots are periodic when written as continued fractions and can be written in the form:
-- √N = a0 + 	
--              1
--   	a1 + 	
--                1
--   	  	a2 + 	
--                  1
--   	  	  	a3 + ...
-- 
-- For example, let us consider √23:
-- √23 = 4 + √23 — 4 = 4 +  	
--          1
-- 	 = 4 +  	
--          1
--   	
--          1
-- √23—4
-- 	  	1 +  	
-- √23 – 3
-- 7
-- 
-- If we continue we would get the following expansion:
-- √23 = 4 + 	
-- 1
--   	1 + 	
-- 1
--   	  	3 + 	
-- 1
--   	  	  	1 + 	
-- 1
--   	  	  	  	8 + ...
-- 
-- The process can be summarised as follows:
-- a0 = 4, 	  	
-- 1
-- √23—4
-- 	 =  	
-- √23+4
-- 7
-- 	 = 1 +  	
-- √23—3
-- 7
-- a1 = 1, 	  	
-- 7
-- √23—3
-- 	 =  	
-- 7(√23+3)
-- 14
-- 	 = 3 +  	
-- √23—3
-- 2
-- a2 = 3, 	  	
-- 2
-- √23—3
-- 	 =  	
-- 2(√23+3)
-- 14
-- 	 = 1 +  	
-- √23—4
-- 7
-- a3 = 1, 	  	
-- 7
-- √23—4
-- 	 =  	
-- 7(√23+4)
-- 7
-- 	 = 8 +  	√23—4
-- a4 = 8, 	  	
-- 1
-- √23—4
-- 	 =  	
-- √23+4
-- 7
-- 	 = 1 +  	
-- √23—3
-- 7
-- a5 = 1, 	  	
-- 7
-- √23—3
-- 	 =  	
-- 7(√23+3)
-- 14
-- 	 = 3 +  	
-- √23—3
-- 2
-- a6 = 3, 	  	
-- 2
-- √23—3
-- 	 =  	
-- 2(√23+3)
-- 14
-- 	 = 1 +  	
-- √23—4
-- 7
-- a7 = 1, 	  	
-- 7
-- √23—4
-- 	 =  	
-- 7(√23+4)
-- 7
-- 	 = 8 +  	√23—4
-- 
-- It can be seen that the sequence is repeating. For conciseness, we use the notation √23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats indefinitely.
-- 
-- The first ten continued fraction representations of (irrational) square roots are:
-- 
-- √2=[1;(2)], period=1
-- √3=[1;(1,2)], period=2
-- √5=[2;(4)], period=1
-- √6=[2;(2,4)], period=2
-- √7=[2;(1,1,1,4)], period=4
-- √8=[2;(1,4)], period=2
-- √10=[3;(6)], period=1
-- √11=[3;(3,6)], period=2
-- √12= [3;(2,6)], period=2
-- √13=[3;(1,1,1,1,6)], period=5
-- 
-- Exactly four continued fractions, for N ≤ 13, have an odd period.
-- 
-- How many continued fractions for N ≤ 10000 have an odd period?

import Control.Monad
import Data.List
import Data.Maybe

f :: Integer -> Integer -> Integer -> [(Integer,(Integer,Integer,Integer))]
f a b c =
    let m = a - (b^2)
        fa = fromInteger a :: Float
        fb = fromInteger b :: Float
        fc = fromInteger c :: Float
        fm = fromInteger m :: Float
        x = floor $ (fc * ((sqrt fa) - fb))/ (fm)
        h = c
        i = -1 *  ( c*b + x*m )
        j = m
    in 
        if ( x^2 == a ) 
           then (0,(0,0,0)):f a b c
           else if (m `rem` h /= 0) 
                then error "Bad division" 
                else (x,(a,b,c)):f a (i `div` h) (j `div` h)
 
lengthPeriod x =
    let s = f x 0 x 
        sig = head $ tail s
        index = fromJust $ elemIndex sig $( tail $ tail s )
    in if head s == (0,(0,0,0))
          then 0
          else index + 1

main = do
    print $ length $ filter (\x -> ((lengthPeriod x) `rem` 2) == 1) [2..10000]
    -- forM_ [2..13] $ \n -> do 
    --     print $ lengthPeriod n
    --     -- print $ tail $  take 10 $ map fst $ f n 0 n