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Development of Algorithmic Constructions

01:12:23 17.Apr 2024

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Polynom = x^2+114x-503

1. Algorithm

a:=1;
b:=114;
c:=-503;


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+114x-503

f(0)=503
f(1)=97
f(2)=271
f(3)=19
f(4)=31
f(5)=23
f(6)=7
f(7)=43
f(8)=11
f(9)=151
f(10)=67
f(11)=109
f(12)=1009
f(13)=41
f(14)=1289
f(15)=179
f(16)=83
f(17)=431
f(18)=1873
f(19)=1
f(20)=311
f(21)=53
f(22)=131
f(23)=331
f(24)=1
f(25)=743
f(26)=3137
f(27)=59
f(28)=1
f(29)=911
f(30)=347
f(31)=499
f(32)=379
f(33)=1087
f(34)=647
f(35)=1
f(36)=1
f(37)=1
f(38)=5273
f(39)=683
f(40)=5657
f(41)=1
f(42)=263
f(43)=71
f(44)=6449
f(45)=1663
f(46)=6857
f(47)=883
f(48)=1039
f(49)=1871
f(50)=1
f(51)=1
f(52)=739
f(53)=2087
f(54)=1
f(55)=157
f(56)=127
f(57)=2311
f(58)=9473
f(59)=1213
f(60)=523
f(61)=2543
f(62)=1487
f(63)=1
f(64)=10889
f(65)=1
f(66)=367
f(67)=1453
f(68)=383
f(69)=433
f(70)=12377
f(71)=1579
f(72)=12889
f(73)=173
f(74)=1
f(75)=1709
f(76)=181
f(77)=1
f(78)=353
f(79)=1
f(80)=15017
f(81)=3823
f(82)=15569
f(83)=283
f(84)=1
f(85)=373
f(86)=1
f(87)=193
f(88)=751
f(89)=4391
f(90)=2551
f(91)=2269
f(92)=971
f(93)=1
f(94)=443
f(95)=1
f(96)=1787
f(97)=1
f(98)=1
f(99)=1

b) Substitution of the polynom
The polynom f(x)=x^2+114x-503  could be written as f(y)= y^2-3752 with x=y-57

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+57
f'(x)>2x+113

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 : 503, 97, 271, 19, 31, 23, 7, 43, 11, 151, 67, 109, 1009, 41, 1289, 179, 83, 431, 1873, 1,
Found in Database : 503, 97, 271, 19, 31, 23, 7, 43, 11, 151, 67, 109, 1009, 41, 1289, 179, 83, 431, 1873, 311, 53, 131, 331, 743, 3137, 59, 911, 347, 499, 379, 1087, 647, 5273, 683,
Found in Database : 7, 11, 19, 23, 31, 41, 43, 53, 59, 67, 71, 83, 97, 109, 127, 131,