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Development of Algorithmic Constructions

10:49:21 16.Apr 2024

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Polynom = x^2+103

1. Algorithm

a:=1;
b:=0;
c:=103;


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+103

f(0)=103
f(1)=13
f(2)=107
f(3)=7
f(4)=17
f(5)=1
f(6)=139
f(7)=19
f(8)=167
f(9)=23
f(10)=29
f(11)=1
f(12)=1
f(13)=1
f(14)=1
f(15)=41
f(16)=359
f(17)=1
f(18)=61
f(19)=1
f(20)=503
f(21)=1
f(22)=587
f(23)=79
f(24)=97
f(25)=1
f(26)=1
f(27)=1
f(28)=887
f(29)=59
f(30)=1
f(31)=1
f(32)=1
f(33)=149
f(34)=1259
f(35)=83
f(36)=1399
f(37)=1
f(38)=1
f(39)=1
f(40)=131
f(41)=223
f(42)=1867
f(43)=1
f(44)=2039
f(45)=1
f(46)=317
f(47)=1
f(48)=1
f(49)=313
f(50)=137
f(51)=1
f(52)=401
f(53)=1
f(54)=3019
f(55)=1
f(56)=1
f(57)=419
f(58)=3467
f(59)=1
f(60)=1
f(61)=239
f(62)=3947
f(63)=509
f(64)=1
f(65)=541
f(66)=1
f(67)=1
f(68)=163
f(69)=1
f(70)=5003
f(71)=643
f(72)=311
f(73)=1
f(74)=797
f(75)=179
f(76)=5879
f(77)=1
f(78)=269
f(79)=1
f(80)=929
f(81)=1
f(82)=6827
f(83)=1
f(84)=7159
f(85)=229
f(86)=7499
f(87)=1
f(88)=1
f(89)=1
f(90)=631
f(91)=1
f(92)=659
f(93)=547
f(94)=1277
f(95)=1
f(96)=9319
f(97)=1
f(98)=571
f(99)=619

b) Substitution of the polynom
The polynom f(x)=x^2+103  could be written as f(y)= y^2+103 with x=y+0

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+0
f'(x)>2x-1  with x > 10

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 : 103, 13, 107, 7, 17, 1, 139, 19, 167, 23, 29, 1, 1, 1, 1, 41, 359, 1, 61, 1,
Found in Database : 103, 13, 107, 7, 17, 139, 19, 167, 23, 29, 41, 359, 61, 503, 587, 79, 97, 887, 59, 149, 1259, 83, 1399,
Found in Database : 7, 13, 17, 19, 23, 29, 41, 59, 61, 79, 83, 97, 103, 107, 131, 137, 139, 149,