Problem Statement

Lucas-Lehmer Test: for

p

{\displaystyle p}

an odd prime, the Mersenne number

2

p

− 1

{\displaystyle 2^{p}-1}

is prime if and only if

2

p

− 1

{\displaystyle 2^{p}-1}

divides

S ( p − 1 )

{\displaystyle S(p-1)}

where

S ( n + 1 )

( S ( n )

)

2

− 2

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

, and

S ( 1 )

4

{\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:

m = 15
n = 1
For j = 2 To m
    If j = 2 Then
        s = 0
    Else
        s = 4
    EndIf
    n = (n + 1) * 2 - 1
    For i = 1 To j - 2
        s = (s * s - 2) % n
    Next i
    If s = 0 Then Print "M";j
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