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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+136x-1249
f(0)=1249
f(1)=139
f(2)=7
f(3)=13
f(4)=53
f(5)=17
f(6)=397
f(7)=31
f(8)=97
f(9)=1
f(10)=211
f(11)=23
f(12)=1
f(13)=43
f(14)=37
f(15)=127
f(16)=1
f(17)=1
f(18)=1523
f(19)=1
f(20)=1871
f(21)=1
f(22)=131
f(23)=1
f(24)=2591
f(25)=347
f(26)=2963
f(27)=197
f(28)=3343
f(29)=1
f(30)=41
f(31)=491
f(32)=4127
f(33)=541
f(34)=1
f(35)=1
f(36)=4943
f(37)=1
f(38)=173
f(39)=1
f(40)=5791
f(41)=751
f(42)=479
f(43)=1
f(44)=953
f(45)=431
f(46)=419
f(47)=919
f(48)=7583
f(49)=977
f(50)=83
f(51)=1
f(52)=8527
f(53)=137
f(54)=9011
f(55)=89
f(56)=1
f(57)=1
f(58)=1429
f(59)=641
f(60)=457
f(61)=673
f(62)=11027
f(63)=1
f(64)=11551
f(65)=1
f(66)=281
f(67)=193
f(68)=971
f(69)=1
f(70)=13171
f(71)=1
f(72)=1
f(73)=103
f(74)=461
f(75)=911
f(76)=167
f(77)=947
f(78)=15443
f(79)=1
f(80)=1
f(81)=157
f(82)=1279
f(83)=1
f(84)=17231
f(85)=1
f(86)=2549
f(87)=2269
f(88)=499
f(89)=2347
f(90)=1123
f(91)=1213
f(92)=19727
f(93)=179
f(94)=1567
f(95)=199
f(96)=21023
f(97)=1
f(98)=21683
f(99)=1
b) Substitution of the polynom
The polynom f(x)=x^2+136x-1249 could be written as f(y)= y^2-5873 with x=y-68
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+68
f'(x)>2x+135