Development of |
|
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+92x-997
f(0)=997
f(1)=113
f(2)=809
f(3)=89
f(4)=613
f(5)=1
f(6)=409
f(7)=19
f(8)=197
f(9)=11
f(10)=23
f(11)=17
f(12)=251
f(13)=1
f(14)=487
f(15)=1
f(16)=43
f(17)=107
f(18)=983
f(19)=139
f(20)=1
f(21)=1
f(22)=1511
f(23)=103
f(24)=1787
f(25)=241
f(26)=109
f(27)=277
f(28)=1
f(29)=157
f(30)=2663
f(31)=1
f(32)=2971
f(33)=1
f(34)=173
f(35)=431
f(36)=1
f(37)=59
f(38)=3943
f(39)=257
f(40)=4283
f(41)=557
f(42)=421
f(43)=601
f(44)=4987
f(45)=1
f(46)=5351
f(47)=1
f(48)=97
f(49)=739
f(50)=359
f(51)=787
f(52)=6491
f(53)=1
f(54)=71
f(55)=443
f(56)=317
f(57)=937
f(58)=7703
f(59)=1
f(60)=8123
f(61)=521
f(62)=503
f(63)=137
f(64)=1
f(65)=1151
f(66)=9431
f(67)=1
f(68)=9883
f(69)=79
f(70)=10343
f(71)=661
f(72)=569
f(73)=1381
f(74)=11287
f(75)=131
f(76)=149
f(77)=751
f(78)=12263
f(79)=1
f(80)=12763
f(81)=1627
f(82)=577
f(83)=1
f(84)=811
f(85)=439
f(86)=1301
f(87)=911
f(88)=14843
f(89)=1889
f(90)=15383
f(91)=1
f(92)=179
f(93)=1013
f(94)=16487
f(95)=1
f(96)=1
f(97)=1
f(98)=17623
f(99)=2239
b) Substitution of the polynom
The polynom f(x)=x^2+92x-997 could be written as f(y)= y^2-3113 with x=y-46
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+46
f'(x)>2x+91