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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+64x-2689
f(0)=2689
f(1)=41
f(2)=2557
f(3)=311
f(4)=2417
f(5)=293
f(6)=2269
f(7)=137
f(8)=2113
f(9)=127
f(10)=1949
f(11)=233
f(12)=1777
f(13)=211
f(14)=1597
f(15)=47
f(16)=1409
f(17)=1
f(18)=1213
f(19)=139
f(20)=1009
f(21)=113
f(22)=797
f(23)=43
f(24)=577
f(25)=29
f(26)=349
f(27)=1
f(28)=1
f(29)=1
f(30)=131
f(31)=1
f(32)=383
f(33)=1
f(34)=643
f(35)=97
f(36)=911
f(37)=1
f(38)=1187
f(39)=83
f(40)=1471
f(41)=101
f(42)=1
f(43)=239
f(44)=2063
f(45)=277
f(46)=2371
f(47)=79
f(48)=2687
f(49)=89
f(50)=3011
f(51)=397
f(52)=3343
f(53)=439
f(54)=1
f(55)=241
f(56)=1
f(57)=263
f(58)=107
f(59)=571
f(60)=4751
f(61)=617
f(62)=109
f(63)=1
f(64)=5503
f(65)=1
f(66)=1
f(67)=761
f(68)=6287
f(69)=811
f(70)=6691
f(71)=431
f(72)=7103
f(73)=457
f(74)=7523
f(75)=967
f(76)=7951
f(77)=1021
f(78)=8387
f(79)=269
f(80)=8831
f(81)=283
f(82)=9283
f(83)=1
f(84)=9743
f(85)=1
f(86)=10211
f(87)=653
f(88)=10687
f(89)=683
f(90)=11171
f(91)=1427
f(92)=1
f(93)=1489
f(94)=12163
f(95)=1
f(96)=12671
f(97)=1
f(98)=13187
f(99)=1
b) Substitution of the polynom
The polynom f(x)=x^2+64x-2689 could be written as f(y)= y^2-3713 with x=y-32
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+32
f'(x)>2x+63