Prime Numbers

3.2 Sieving 125
noticed that it was essentially sufficient to sieve just the odd numbers up to
N when searching for primes.
To sieve the sequence f(1),f(2),...,f(N), we initialize with ones an array
corresponding to the numbers 1, 2,...,N. An important observation is that
if p is prime and a satisfies f(a) ≡ 0(modp), then f(a + kp) ≡ 0(modp)
for every integer k. Of course, there may be as many as degf such solutions
a, and hence just as many distinct arithmetic progressions {a + kp} for each
sieving prime p.
Let us illustrate with the polynomial f(x) =x 2 + 1. We wish to find the
primes of the form x 2 +1 for x an integer, 1 ≤ x ≤ N. For each prime p ≤ N,
solve the congruence x 2 +1 ≡ 0(modp) (see Section 2.3.2). When p ≡ 1
(mod 4), there are two solutions, when p ≡ 3 (mod 4), there are no solutions,
and when p = 2 there is exactly one solution. For each prime p and solution
a (that is, a 2 +1≡ 0(modp) and1≤ a