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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^2-270x+653
f(0)=653
f(1)=3
f(2)=13
f(3)=37
f(4)=137
f(5)=7
f(6)=19
f(7)=11
f(8)=1
f(9)=53
f(10)=59
f(11)=61
f(12)=349
f(13)=1
f(14)=977
f(15)=1
f(16)=379
f(17)=1
f(18)=353
f(19)=1
f(20)=23
f(21)=1
f(22)=1601
f(23)=419
f(24)=89
f(25)=1
f(26)=271
f(27)=211
f(28)=157
f(29)=1
f(30)=6547
f(31)=563
f(32)=1
f(33)=1
f(34)=1
f(35)=631
f(36)=409
f(37)=83
f(38)=907
f(39)=2089
f(40)=1
f(41)=1
f(42)=8923
f(43)=1
f(44)=163
f(45)=1
f(46)=3217
f(47)=1
f(48)=1429
f(49)=1
f(50)=3449
f(51)=239
f(52)=1187
f(53)=113
f(54)=1
f(55)=1
f(56)=1259
f(57)=359
f(58)=3881
f(59)=983
f(60)=919
f(61)=1
f(62)=1
f(63)=1
f(64)=4177
f(65)=1
f(66)=557
f(67)=1
f(68)=1
f(69)=1
f(70)=1483
f(71)=1123
f(72)=223
f(73)=1
f(74)=1
f(75)=499
f(76)=1
f(77)=1
f(78)=14323
f(79)=401
f(80)=373
f(81)=229
f(82)=1
f(83)=1
f(84)=1361
f(85)=1
f(86)=389
f(87)=347
f(88)=569
f(89)=1
f(90)=2221
f(91)=1303
f(92)=1747
f(93)=1
f(94)=5297
f(95)=1
f(96)=2293
f(97)=1
f(98)=491
f(99)=313
b) Substitution of the polynom
The polynom f(x)=x^2-270x+653 could be written as f(y)= y^2-17572 with x=y+135
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-135
f'(x)>2x-271 with x > 133