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Development of Algorithmic Constructions

09:26:58 20.Apr 2024

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Polynom = x^2-12x+5

1. Algorithm

a:=1;
b:=-12;
c:=5;


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^2-12x+5

f(0)=5
f(1)=3
f(2)=1
f(3)=11
f(4)=1
f(5)=1
f(6)=31
f(7)=1
f(8)=1
f(9)=1
f(10)=1
f(11)=1
f(12)=1
f(13)=1
f(14)=1
f(15)=1
f(16)=23
f(17)=1
f(18)=113
f(19)=1
f(20)=1
f(21)=97
f(22)=1
f(23)=43
f(24)=293
f(25)=1
f(26)=41
f(27)=1
f(28)=151
f(29)=83
f(30)=109
f(31)=1
f(32)=1
f(33)=349
f(34)=251
f(35)=1
f(36)=79
f(37)=1
f(38)=331
f(39)=1
f(40)=1
f(41)=199
f(42)=1
f(43)=223
f(44)=157
f(45)=149
f(46)=523
f(47)=1
f(48)=1733
f(49)=101
f(50)=127
f(51)=997
f(52)=139
f(53)=1
f(54)=2273
f(55)=1
f(56)=823
f(57)=257
f(58)=1
f(59)=463
f(60)=577
f(61)=499
f(62)=1
f(63)=1609
f(64)=1
f(65)=1
f(66)=1
f(67)=1
f(68)=1
f(69)=179
f(70)=271
f(71)=233
f(72)=173
f(73)=743
f(74)=1531
f(75)=1
f(76)=541
f(77)=167
f(78)=5153
f(79)=883
f(80)=1
f(81)=2797
f(82)=383
f(83)=983
f(84)=6053
f(85)=1
f(86)=193
f(87)=653
f(88)=1
f(89)=1
f(90)=281
f(91)=1
f(92)=491
f(93)=3769
f(94)=857
f(95)=263
f(96)=8069
f(97)=1
f(98)=937
f(99)=1

b) Substitution of the polynom
The polynom f(x)=x^2-12x+5  could be written as f(y)= y^2-31 with x=y+6

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-6
f'(x)>2x-13  with x > 6

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 : 5, 3, 1, 11, 1, 1, 31, 1, 1, 1, 1, 1, 1, 1, 1, 1, 23, 1, 113, 1,
Found in Database : 5, 3, 11, 31, 23, 113, 97, 43, 293, 41, 151, 83, 109, 349, 251, 79, 331,
Found in Database : 3, 5, 11, 23, 31, 41, 43, 79, 83, 97, 101, 109, 113, 127, 139, 149,