---

Development of Algorithmic Constructions

00:33:59 20.Apr 2024

---

Polynom = x^2+128x-1033

1. Algorithm

a:=1;
b:=128;
c:=-1033;


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+128x-1033

f(0)=1033
f(1)=113
f(2)=773
f(3)=5
f(4)=101
f(5)=23
f(6)=229
f(7)=11
f(8)=1
f(9)=1
f(10)=347
f(11)=31
f(12)=647
f(13)=1
f(14)=191
f(15)=139
f(16)=41
f(17)=179
f(18)=29
f(19)=1
f(20)=47
f(21)=131
f(22)=2267
f(23)=61
f(24)=523
f(25)=349
f(26)=2971
f(27)=197
f(28)=1
f(29)=1
f(30)=337
f(31)=487
f(32)=67
f(33)=107
f(34)=1
f(35)=73
f(36)=4871
f(37)=317
f(38)=211
f(39)=137
f(40)=1
f(41)=1
f(42)=1
f(43)=79
f(44)=1307
f(45)=1
f(46)=6971
f(47)=1
f(48)=1483
f(49)=1
f(50)=7867
f(51)=1
f(52)=757
f(53)=1
f(54)=1759
f(55)=1129
f(56)=127
f(57)=1
f(58)=1951
f(59)=1
f(60)=10247
f(61)=1
f(62)=977
f(63)=1
f(64)=2251
f(65)=1439
f(66)=149
f(67)=1
f(68)=2459
f(69)=157
f(70)=1
f(71)=1637
f(72)=13367
f(73)=1
f(74)=1
f(75)=887
f(76)=499
f(77)=461
f(78)=97
f(79)=383
f(80)=15607
f(81)=1987
f(82)=16187
f(83)=103
f(84)=1
f(85)=1
f(86)=599
f(87)=1
f(88)=719
f(89)=457
f(90)=18587
f(91)=1181
f(92)=19207
f(93)=1
f(94)=3967
f(95)=1
f(96)=1861
f(97)=1
f(98)=1
f(99)=1

b) Substitution of the polynom
The polynom f(x)=x^2+128x-1033  could be written as f(y)= y^2-5129 with x=y-64

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+64
f'(x)>2x+127

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 : 1033, 113, 773, 5, 101, 23, 229, 11, 1, 1, 347, 31, 647, 1, 191, 139, 41, 179, 29, 1,
Found in Database : 1033, 113, 773, 5, 101, 23, 229, 11, 347, 31, 647, 191, 139, 41, 179, 29, 47, 131, 2267, 61, 523, 349, 2971, 197, 337, 487, 67, 107, 73, 4871, 317, 211, 137,
Found in Database : 5, 11, 23, 29, 31, 41, 47, 61, 67, 73, 79, 97, 101, 103, 107, 113, 127, 131, 137, 139, 149,