/* 

  Number theory problem in 2012 CMO in Picat.

  http://community.wolfram.com/groups/-/m/t/793922
  """
  Given two positive integers a and b. The two numbers satisfy the following conditions: 
  a−b is a prime number p and a×b is a perfect square n^2 . 
  Find the smallest value of a no less than 2012.
  """

  This Picat model was created by Hakan Kjellerstrand, hakank@gmail.com
  See also my Picat page: http://www.hakank.org/picat/

*/


% import util.
% import sat.
import cp.
% import math.


main => go.

% sat: 2.8s
% cp: 0.07s
go =>
  nolog,
  Primes = primes(2012),
  A :: 2012..10000,
  B :: 2..2012,
  P :: Primes,
  N :: 2..2012,

  A #>= B,
  P #= A-B,
  A*B #= N*N,

  Vars = [P,B,N],
  solve($[min(A),up,report(printf("A: %d\n", A))],Vars),

  println(a=A),
  println(b=B),
  println(p=P),
  % println(z=Z),
  println(n=N),
    
  
  nl.

% sat: 1.5s
% cp: 0.065s
go2 =>
  nolog,
  Max = 10000,
  Primes = primes(2012),
  member(A,2012..Max),
  % println(testing=A),
  B :: 2..2012,
  P :: Primes,
  N :: 2..2012,

  A #>= B,
  P #= A-B,
  A*B #= N*N,

  Vars = [P,B,N],
  solve($[min(A),up,report(printf("A: %d\n", A))],Vars),

  println(a=A),
  println(b=B),
  println(p=P),
  println(n=N),
    
  
  nl.