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Development of Algorithmic Constructions

07:25:23 18.Apr 2024

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Polynom = x^2+124x-3

1. Algorithm

a:=1;
b:=124;
c:=-3;


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+124x-3

f(0)=3
f(1)=61
f(2)=83
f(3)=7
f(4)=509
f(5)=107
f(6)=37
f(7)=457
f(8)=13
f(9)=199
f(10)=191
f(11)=19
f(12)=181
f(13)=127
f(14)=643
f(15)=347
f(16)=2237
f(17)=1
f(18)=23
f(19)=59
f(20)=137
f(21)=1
f(22)=3209
f(23)=563
f(24)=1
f(25)=1861
f(26)=433
f(27)=97
f(28)=4253
f(29)=739
f(30)=1
f(31)=1
f(32)=1663
f(33)=863
f(34)=1
f(35)=103
f(36)=101
f(37)=229
f(38)=293
f(39)=353
f(40)=79
f(41)=1
f(42)=1
f(43)=1
f(44)=821
f(45)=1
f(46)=7817
f(47)=1
f(48)=131
f(49)=223
f(50)=1
f(51)=1487
f(52)=1307
f(53)=521
f(54)=3203
f(55)=1
f(56)=3359
f(57)=1
f(58)=173
f(59)=257
f(60)=283
f(61)=5641
f(62)=1
f(63)=151
f(64)=523
f(65)=89
f(66)=1
f(67)=6397
f(68)=1
f(69)=317
f(70)=13577
f(71)=769
f(72)=4703
f(73)=1
f(74)=1
f(75)=829
f(76)=167
f(77)=2579
f(78)=1
f(79)=8017
f(80)=1
f(81)=2767
f(82)=16889
f(83)=409
f(84)=647
f(85)=1
f(86)=463
f(87)=1
f(88)=811
f(89)=1
f(90)=1
f(91)=9781
f(92)=179
f(93)=1
f(94)=2927
f(95)=3467
f(96)=7039
f(97)=1531
f(98)=2417
f(99)=1

b) Substitution of the polynom
The polynom f(x)=x^2+124x-3  could be written as f(y)= y^2-3847 with x=y-62

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+62
f'(x)>2x+123

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 : 3, 61, 83, 7, 509, 107, 37, 457, 13, 199, 191, 19, 181, 127, 643, 347, 2237, 1, 23, 59,
Found in Database : 3, 61, 83, 7, 509, 107, 37, 457, 13, 199, 191, 19, 181, 127, 643, 347, 2237, 23, 59, 137, 3209, 563, 1861, 433, 97, 4253, 739, 1663, 863, 103, 101, 229, 293, 353,
Found in Database : 3, 7, 13, 19, 23, 37, 59, 61, 79, 83, 89, 97, 101, 103, 107, 127, 131, 137,