pb 66 solution

066.hs

@@ -0,0 +1,46 @@
+-- Consider quadratic Diophantine equations of the form:
+-- 
+-- x^2 – Dy^2 = 1
+-- 
+-- For example, when D=13, the minimal solution in x is 6492 – 13×1802 = 1.
+-- 
+-- It can be assumed that there are no solutions in positive integers when D is square.
+-- 
+-- By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:
+-- 
+--         32 – 2×22 = 1
+--         22 – 3×12 = 1
+--         92 – 5×42 = 1
+--         52 – 6×22 = 1
+--         82 – 7×32 = 1
+-- 
+-- Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5.
+-- 
+-- Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.
+
+import Control.Monad
+import Data.List
+import Data.Maybe
+
+isqrt = floor . sqrt . fromIntegral
+
+isSquare d = (isqrt d)^2 == d
+
+hasSolution d x =
+    let y = isqrt (( x^2 - 1 ) `div` d)
+    in x^2 - (y^2)*d == 1
+
+phead [] = -2 
+phead (x:xs) = x
+
+sol d = if isSquare d
+           then -1 
+           else phead $ filter (hasSolution d) [2..1000000]
+
+main = let 
+            lst = map sol [2..1000] 
+            maxelem = foldl' (\x y -> if x>y then x else y) 0 lst
+       in do
+           forM_ [2..1000] (\d -> do putStr $ (show d) ++ " -> "; print $ lst !! d)
+           print maxelem
+           print $ elemIndex maxelem lst