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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+10x+821
f(0)=821
f(1)=13
f(2)=5
f(3)=43
f(4)=877
f(5)=7
f(6)=131
f(7)=47
f(8)=193
f(9)=31
f(10)=1021
f(11)=263
f(12)=1
f(13)=1
f(14)=89
f(15)=23
f(16)=1237
f(17)=1
f(18)=53
f(19)=1
f(20)=29
f(21)=1
f(22)=61
f(23)=79
f(24)=1637
f(25)=1
f(26)=251
f(27)=1
f(28)=1
f(29)=1
f(30)=1
f(31)=523
f(32)=433
f(33)=1
f(34)=331
f(35)=599
f(36)=2477
f(37)=1
f(38)=1
f(39)=683
f(40)=1
f(41)=1
f(42)=601
f(43)=1
f(44)=139
f(45)=103
f(46)=1
f(47)=1
f(48)=1
f(49)=1
f(50)=3821
f(51)=983
f(52)=809
f(53)=1
f(54)=1
f(55)=157
f(56)=4517
f(57)=1
f(58)=953
f(59)=1223
f(60)=5021
f(61)=1
f(62)=151
f(63)=271
f(64)=5557
f(65)=1
f(66)=449
f(67)=1
f(68)=1
f(69)=1
f(70)=6421
f(71)=1
f(72)=269
f(73)=1
f(74)=227
f(75)=257
f(76)=1051
f(77)=1
f(78)=1
f(79)=1
f(80)=617
f(81)=1
f(82)=239
f(83)=1
f(84)=379
f(85)=1
f(86)=313
f(87)=463
f(88)=1889
f(89)=1
f(90)=1
f(91)=2503
f(92)=1
f(93)=1
f(94)=10597
f(95)=2699
f(96)=1571
f(97)=1
f(98)=2281
f(99)=2903
b) Substitution of the polynom
The polynom f(x)=x^2+10x+821 could be written as f(y)= y^2+796 with x=y-5
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+5
f'(x)>2x+9