A second proof for the LucasLehmer Test
The LucasLehmer test is a test for Mersenne Numbers 2^n1.
2^n1 is prime if and only if 2^n1 divides S(4, n2).
Here S(4, n) = S(4, n1)^22
starting with S(4, 0) = 4
now suppose we replace S(4, 0) = 4 with S(x, 0) = x.
We get the following polynomial sequence
S(x, 0) = x
S(x, 1) = x^22
S(x, 2) = x^44*x^2+2
S(x, 3) = x^88*x^6+20*x^416*x^2+2
S(x, 4) = x^1616*x^14+104*x^12352*x^10+660*x^8672*x^6+336*x^464*x^22
...
Now we see that the discriminant D of S(x, n) = 2^r where r = (n+1)*2^n1
Since the exponent r is odd, each prime factor q dividing S(x, n) has the form k*2^(n+1)+1.
Now assume 2^n1 divides S(4, n2).
This shows that each prime factor of 2^n1 must have the form k*2^(n1)+1. Since k*2^(n1)+1 > sqrt(2^n1) is always true for k > 0, there is no prime factor less than or equal to sqrt(2^n1) with that form. By trial division test if c is composite, there exists a prime p < sqrt(c) that divides c. Therefore, 2^n1 must be prime. Please feel free to comment, suggest, improve or ask on this. Thanks!!!
Last fiddled with by carpetpool on 20170730 at 00:31
