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Development of Algorithmic Constructions

01:46:25 20.Apr 2024

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Polynom = x^2-250x-3

1. Algorithm

a:=1;
b:=-250;
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^2-250x-3

f(0)=3
f(1)=7
f(2)=499
f(3)=31
f(4)=47
f(5)=307
f(6)=163
f(7)=71
f(8)=277
f(9)=181
f(10)=89
f(11)=1
f(12)=953
f(13)=257
f(14)=3307
f(15)=1
f(16)=1249
f(17)=991
f(18)=199
f(19)=61
f(20)=4603
f(21)=401
f(22)=239
f(23)=653
f(24)=67
f(25)=1
f(26)=5827
f(27)=251
f(28)=691
f(29)=229
f(30)=1
f(31)=283
f(32)=997
f(33)=1
f(34)=79
f(35)=941
f(36)=367
f(37)=73
f(38)=8059
f(39)=1
f(40)=2801
f(41)=2143
f(42)=971
f(43)=53
f(44)=9067
f(45)=769
f(46)=149
f(47)=1193
f(48)=1
f(49)=821
f(50)=1429
f(51)=1
f(52)=3433
f(53)=373
f(54)=3529
f(55)=1
f(56)=10867
f(57)=131
f(58)=1
f(59)=1409
f(60)=1
f(61)=1
f(62)=1
f(63)=491
f(64)=1
f(65)=97
f(66)=4049
f(67)=1
f(68)=12379
f(69)=347
f(70)=4201
f(71)=227
f(72)=4273
f(73)=359
f(74)=1861
f(75)=547
f(76)=4409
f(77)=3331
f(78)=1
f(79)=563
f(80)=223
f(81)=1
f(82)=1531
f(83)=1733
f(84)=4649
f(85)=167
f(86)=14107
f(87)=197
f(88)=1
f(89)=3583
f(90)=4801
f(91)=1
f(92)=1
f(93)=1217
f(94)=4889
f(95)=263
f(96)=1
f(97)=1237
f(98)=317
f(99)=1

b) Substitution of the polynom
The polynom f(x)=x^2-250x-3  could be written as f(y)= y^2-15628 with x=y+125

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-125
f'(x)>2x-251  with x > 125

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, 7, 499, 31, 47, 307, 163, 71, 277, 181, 89, 1, 953, 257, 3307, 1, 1249, 991, 199, 61,
Found in Database : 3, 7, 499, 31, 47, 307, 163, 71, 277, 181, 89, 953, 257, 3307, 1249, 991, 199, 61, 4603, 401, 239, 653, 67, 5827, 251, 691, 229, 283, 997, 79, 941, 367, 73, 8059,
Found in Database : 3, 7, 31, 47, 53, 61, 67, 71, 73, 79, 89, 97, 131, 149,