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liste_max:=100000; sieving:=proc (stelle, p) begin while (stelle<=liste_max) do erg:=liste[stelle]; while(erg mod p=0) do // Divison of the stored f(x) by the prime erg:=erg /p; end_while; liste[stelle]:=erg; stelle:=stelle+p; end_while; end_proc; // Calculation of the values of the polynom for x from 0 to liste_max for x from 0 to liste_max do p:=abs (a*x^2+b*x+c); while (p mod 2=0) p:=p/2; liste [x]:=p; end_for; for x from 0 to liste_max do p:=liste[x]; if (p>1) then // Printing the Primes print (x, p); // 1. Sieving sieving (x+p, p); t:=(-x-b/a) mod p;If you are interested in some better algorithms have a look at quadr_Sieb_x^2+1.php.
if t=0 then t:=p; end_if; // 2. Sieving sieving (t, p); end_if; end_for;
2. Mathematical background
Lemma: If p | f(x) then also p | f(x+p) and p | f(-x-b/a) a) p | f(x) <=> ax^2 + bx + c = 0 mod p p | f(x+p) <=> a(x+p)^2 + b(x+p) + c = 0 mod p <=> ax^2 + 2axp + ap^2 + bx + bp + c = 0 mod p <=> ax^2 + bx + c = 0 mod p Thus if p | f(x) then p | f(x+p) b) if b = 0 mod a p | f(x) <=> ax^2 + bx + c = 0 mod p p | f(-x-b/a) <=> a(-x-b/a)^2 + b(-x-b/a) + c = 0 mod p <=> ax^2 + 2bx + b^2/a - bx - b^2/a + c = 0 mod p <=> ax^2 + bx + c = 0 mod p Thus if p | f(x) then p | f(-x-b/a)3. Correctness of the algorithm
The proof for this polynom is similar to the proof for the polynom f(x)=x^2-4x+1. a) First terms for the polynom f(x) = x^2-116x+211
f(0)=211
f(1)=3
f(2)=17
f(3)=1
f(4)=79
f(5)=43
f(6)=449
f(7)=23
f(8)=653
f(9)=47
f(10)=283
f(11)=59
f(12)=61
f(13)=1
f(14)=1217
f(15)=163
f(16)=463
f(17)=1
f(18)=1553
f(19)=1
f(20)=1709
f(21)=223
f(22)=619
f(23)=241
f(24)=1997
f(25)=1
f(26)=2129
f(27)=137
f(28)=751
f(29)=1
f(30)=103
f(31)=101
f(32)=2477
f(33)=1
f(34)=859
f(35)=41
f(36)=157
f(37)=113
f(38)=2753
f(39)=349
f(40)=1
f(41)=179
f(42)=2897
f(43)=1
f(44)=2957
f(45)=373
f(46)=1
f(47)=379
f(48)=71
f(49)=1
f(50)=3089
f(51)=97
f(52)=1039
f(53)=1
f(54)=3137
f(55)=131
f(56)=67
f(57)=197
f(58)=1051
f(59)=1
f(60)=1
f(61)=1
f(62)=1
f(63)=1
f(64)=1
f(65)=1
f(66)=1
f(67)=1
f(68)=1
f(69)=1
f(70)=1
f(71)=1
f(72)=1
f(73)=1
f(74)=1
f(75)=1
f(76)=1
f(77)=1
f(78)=1
f(79)=1
f(80)=1
f(81)=1
f(82)=1
f(83)=1
f(84)=1
f(85)=1
f(86)=1
f(87)=1
f(88)=1
f(89)=1
f(90)=1
f(91)=1
f(92)=1
f(93)=1
f(94)=1
f(95)=1
f(96)=1
f(97)=1
f(98)=1
f(99)=1
b) Substitution of the polynom
The polynom f(x)=x^2-116x+211 could be written as f(y)= y^2-3153 with x=y+58
c) Backsubstitution Beside by backsubstitution you get an estimation for the huge of the primes with p | f(x) and p < f(x) f'(y)>(2y-1) with with y=x-58
f'(x)>2x-117 with x > 56