How to resolve the algorithm Lucas-Lehmer test step by step in the Ada programming language

Problem Statement
Step by Step Solution
Sourcecode

Problem Statement

Lucas-Lehmer Test: for

{\displaystyle p}

an odd prime, the Mersenne number

− 1

{\displaystyle 2^{p}-1}

is prime if and only if

− 1

{\displaystyle 2^{p}-1}

divides

S ( p − 1 )

{\displaystyle S(p-1)}

where

S ( n + 1 )

( S ( n )

)

− 2

{\displaystyle S(n+1)=(S(n))^{2}-2}

, and

S ( 1 )

{\displaystyle S(1)=4}

Calculate all Mersenne primes up to the implementation's maximum precision, or the 47th Mersenne prime (whichever comes first).

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Lucas-Lehmer test step by step in the Ada programming language

Source code in the ada programming language

with Ada.Text_Io; use Ada.Text_Io;
with Ada.Integer_Text_Io; use Ada.Integer_Text_Io;

procedure Lucas_Lehmer_Test is
   type Ull is mod 2**64;
   function Mersenne(Item : Integer) return Boolean is
      S : Ull := 4;
      MP : Ull := 2**Item - 1;
   begin
      if Item = 2 then
         return True;
      else
         for I in 3..Item loop
            S := (S * S - 2) mod MP;
         end loop;
         return S = 0;
      end if;
   end Mersenne;
   Upper_Bound : constant Integer := 64;
begin
   Put_Line(" Mersenne primes:");
   for P in 2..Upper_Bound loop
      if Mersenne(P) then
         Put(" M");
         Put(Item => P, Width => 1);
      end if;
   end loop;
end Lucas_Lehmer_Test;