The sieve of Helmes uses the periodical appearance of the prime numbers on the
irreducible polynoms
1. f(x)=x^2+1 all primes mod 4 = 1 appear as factor 2. f(x)=x^2+x+1 all primes mod 6 = 1 appear as factor
All primes without the prime number 2 appear as factor on these three polynoms
I. f(x)=2x^2-1 all primes mod 8 = 1 or 7 appear as factor
II. g(x)=2x^2+1 all primes mod 8 = 1 or 3 appear as factor
III. h(x)=4x^2+1 all primes mod 8 = 1 or 5 appear as factor
Thesis
All Primes (without the 2) appear as factor on these three polynoms.
Proof
Let p be a prime, p>2
Then there is some x element of N and some {f, g, h} such that p divides t(x)
Assume first that p = 4k+1 for some k element N
Then -1 is a square in Fp* and hence there is some a element Fp* such that
a^2= -1.
Put x´:=2^(-1)a element Fp and choose some x element N such that x´= x mod p
then 4x^2 = 4x´^2 = 4*2^(-2)a^2 = a^2 = -1 mod p.
Hence p divides h (x) = 4x^2+1
The remaining primes are of the form 8k+3 or 8k-1.
Assume that p = 8k+3 for some k element N
Then -2 is a square in Fp and so is -1/2.
So there is x element N such that x^2 = -1/2 mod p
Hence p divides g(x) = 2x^2+1.
Assume finally that p = 8k-1
Then 2 is a square in Fp and so is 1/2.
So there is x element N such that x^2 = 1/2 mod p
Hence p divides f(x) = 2x^2-1.
