content: {- 
   - Solution to Project Euler problem 120
   - Copyright (c) Project Nayuki. All rights reserved.
   - 
   - https://www.nayuki.io/page/project-euler-solutions
   - https://github.com/nayuki/Project-Euler-solutions
   -}
  
  
  {- 
   - For a given a, what is the n that maximizes the remainder, and what is the value of this remainder?
   - 
   - Let's simplify one term, mod a^2:
   -   (a+1)^n = 1^n + (n choose 1) a 1^(n-1) + (n choose 2) a^2 1^(n-2) + ...  (by the binomial theorem)
   -           = 1 + an + 0.  (remaining addends are 0 because they have a to the power of 2 or more, mod a^2)
   - Similarly for the other term, mod a^2:
   -   (a-1)^n = (-1)^n + (n choose 1) a (-1)^(n-1) + ...
   -           = (-1)^(n-1) (-1 + an + 0)
   -           = if n is even then (1 - an) else (an - 1).
   - Therefore, adding the two terms:
   -   (a+1)^n + (a-1)^n
   -   = if n is even then 2 else 2an.
   - 
   - We can always make 2an >= 2 by taking n=1, for example. So we can disregard the "n is even" case.
   - Maximizing 2an mod a^2 for n is the same as maximizing 2n mod a for n.
   -   If a is even,  then the maximum achievable value is a - 2 by setting n = a/2 - 1.
   -   Else a is odd, then the maximum achievable value is a - 1 by setting n = (a - 1) / 2.
   - 
   - In conclusion, if a is even, the maximum remainder is a(a-2);
   - otherwise a is odd, the maximum remainder is a(a-1).
   -}
  main = putStrLn (show ans)
  ans = sum [a * (a - (if (mod a 2) == 0 then 2 else 1)) | a <- [3..1000]]
  