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Development of Algorithmic Constructions

16:48:53 20.Apr 2024

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Polynom = x^2+132x+59

1. Algorithm

a:=1;
b:=132;
c:=59;


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 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+132x+59

f(0)=59
f(1)=3
f(2)=109
f(3)=29
f(4)=67
f(5)=31
f(6)=887
f(7)=43
f(8)=131
f(9)=83
f(10)=17
f(11)=1
f(12)=1787
f(13)=1
f(14)=701
f(15)=283
f(16)=809
f(17)=1
f(18)=89
f(19)=61
f(20)=1033
f(21)=409
f(22)=383
f(23)=151
f(24)=3803
f(25)=1
f(26)=463
f(27)=1
f(28)=1
f(29)=197
f(30)=4919
f(31)=71
f(32)=1
f(33)=1
f(34)=1901
f(35)=41
f(36)=1
f(37)=263
f(38)=53
f(39)=1
f(40)=257
f(41)=149
f(42)=139
f(43)=79
f(44)=1
f(45)=1
f(46)=2749
f(47)=353
f(48)=8699
f(49)=1
f(50)=1
f(51)=587
f(52)=3209
f(53)=137
f(54)=10103
f(55)=431
f(56)=3529
f(57)=677
f(58)=1231
f(59)=1
f(60)=11579
f(61)=1
f(62)=1
f(63)=1543
f(64)=4201
f(65)=1
f(66)=13127
f(67)=1
f(68)=157
f(69)=1741
f(70)=4733
f(71)=1
f(72)=14747
f(73)=313
f(74)=5101
f(75)=487
f(76)=1
f(77)=673
f(78)=967
f(79)=1
f(80)=1
f(81)=541
f(82)=5869
f(83)=373
f(84)=167
f(85)=1
f(86)=6269
f(87)=2389
f(88)=6473
f(89)=1
f(90)=691
f(91)=1
f(92)=1
f(93)=1
f(94)=1
f(95)=1
f(96)=1291
f(97)=1
f(98)=1
f(99)=1433

b) Substitution of the polynom
The polynom f(x)=x^2+132x+59  could be written as f(y)= y^2-4297 with x=y-66

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+66
f'(x)>2x+131

4. Infinity of the sequence

5. Sequence of the polynom with 1

6. Sequence of the polynom (only primes)

7. Distribution of the primes

8. Check for existing Integer Sequences by OEIS

Found in Database : 59, 3, 109, 29, 67, 31, 887, 43, 131, 83, 17, 1, 1787, 1, 701, 283, 809, 1, 89, 61,
Found in Database : 59, 3, 109, 29, 67, 31, 887, 43, 131, 83, 17, 1787, 701, 283, 809, 89, 61, 1033, 409, 383, 151, 3803, 463, 197, 4919, 71, 1901, 41, 263, 53,
Found in Database : 3, 17, 29, 31, 41, 43, 53, 59, 61, 67, 71, 79, 83, 89, 109, 131, 137, 139, 149,