improved divStep http://stackoverflow.com/questions/16387032/how-to-write-this-as-a-single-function/16399024#16399024

divStep.lhs

I'm late to the party but here goes:

>   divStep (start, n) = head $ [(x,q) | x <- [start .. isqrt n],
>                                let (q,r)=quotRem n x, r == 0] ++ [(n,1)]
    
>   isqrt n = floor . sqrt . fromIntegral $ n

>   pe3 n | n > 1 = fst . until ((== 1) . snd) divStep $ (2,n)
    
>   factorizing n = takeWhile ((> 1) . fst) . drop 1 . iterate divStep $ (2,n)
    
>   factors n = map fst . factorizing $ n
    
>   isPrime n = factors n == [n] 

as developed and explained here - http://stackoverflow.com/questions/15699820/
  racket-programming-where-am-i-going-wrong/15703327#15703327 . 
It produces the factors of a number, that by construction are guaranteed to be prime, 
so no additional testing is necessary.

You rightfully felt weird in this particular case, because your `get_factors` is 
nothing but a `filter`, as others pointed out.