Project Euler #9 in haskell
As already mentioned there are formulas for enumerating the pythagorean triples, but your solution can be made quite fast by bounding the search space more. You have already taken care of the symmetry in the solutions by letting a be less then b (a &lt;- [1..b]), but note that since a + b + c == 1000, once you know the value of a and b you know that c = 1000 - a - b so you only need to check that one case:

[(a,b,c) | a &lt;- [1..500], b &lt;- [1..a], let c = 1000 - a - b, a^2 + b^2 == c^2]


The above code runs in less then a second on my machine with GHCi.