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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-449
f(0)=449
f(1)=3
f(2)=317
f(3)=31
f(4)=59
f(5)=13
f(6)=29
f(7)=1
f(8)=127
f(9)=1
f(10)=97
f(11)=47
f(12)=463
f(13)=23
f(14)=643
f(15)=1
f(16)=277
f(17)=1
f(18)=79
f(19)=1
f(20)=1231
f(21)=167
f(22)=37
f(23)=1
f(24)=1663
f(25)=1
f(26)=61
f(27)=251
f(28)=709
f(29)=281
f(30)=2371
f(31)=1
f(32)=43
f(33)=1
f(34)=1
f(35)=1
f(36)=137
f(37)=1
f(38)=149
f(39)=223
f(40)=1237
f(41)=241
f(42)=4003
f(43)=173
f(44)=331
f(45)=557
f(46)=53
f(47)=1
f(48)=379
f(49)=1
f(50)=89
f(51)=677
f(52)=1861
f(53)=719
f(54)=5923
f(55)=1
f(56)=6271
f(57)=1
f(58)=1
f(59)=1
f(60)=6991
f(61)=1
f(62)=199
f(63)=1
f(64)=1
f(65)=1
f(66)=1
f(67)=347
f(68)=8527
f(69)=1091
f(70)=229
f(71)=571
f(72)=9343
f(73)=1
f(74)=751
f(75)=1
f(76)=1
f(77)=1301
f(78)=10627
f(79)=113
f(80)=11071
f(81)=353
f(82)=1
f(83)=1
f(84)=521
f(85)=509
f(86)=12451
f(87)=1
f(88)=139
f(89)=823
f(90)=13411
f(91)=569
f(92)=13903
f(93)=1
f(94)=4801
f(95)=1
f(96)=1
f(97)=1
f(98)=15427
f(99)=1
b) Substitution of the polynom
The polynom f(x)=x^2+64x-449 could be written as f(y)= y^2-1473 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