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Development of Algorithmic Constructions

11:05:05 19.Apr 2024

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Polynom = x^2+5

1. Algorithm

a:=1;
b:=0;
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+5

f(0)=5
f(1)=3
f(2)=1
f(3)=7
f(4)=1
f(5)=1
f(6)=41
f(7)=1
f(8)=23
f(9)=43
f(10)=1
f(11)=1
f(12)=149
f(13)=29
f(14)=67
f(15)=1
f(16)=1
f(17)=1
f(18)=47
f(19)=61
f(20)=1
f(21)=223
f(22)=163
f(23)=89
f(24)=83
f(25)=1
f(26)=227
f(27)=367
f(28)=263
f(29)=1
f(30)=181
f(31)=1
f(32)=1
f(33)=547
f(34)=1
f(35)=1
f(36)=1301
f(37)=229
f(38)=1
f(39)=109
f(40)=107
f(41)=281
f(42)=1
f(43)=103
f(44)=647
f(45)=1
f(46)=101
f(47)=1
f(48)=2309
f(49)=401
f(50)=167
f(51)=1303
f(52)=1
f(53)=1
f(54)=127
f(55)=1
f(56)=349
f(57)=1627
f(58)=1123
f(59)=1
f(60)=1
f(61)=1
f(62)=1283
f(63)=1987
f(64)=1367
f(65)=1
f(66)=1
f(67)=1
f(68)=1543
f(69)=2383
f(70)=1
f(71)=1
f(72)=5189
f(73)=1
f(74)=1
f(75)=563
f(76)=1
f(77)=1
f(78)=6089
f(79)=347
f(80)=1
f(81)=1
f(82)=2243
f(83)=383
f(84)=307
f(85)=241
f(86)=2467
f(87)=541
f(88)=1
f(89)=1321
f(90)=1621
f(91)=1381
f(92)=941
f(93)=4327
f(94)=421
f(95)=1
f(96)=9221
f(97)=523
f(98)=3203
f(99)=4903

b) Substitution of the polynom
The polynom f(x)=x^2+5  could be written as f(y)= y^2+5 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 > 2

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, 7, 1, 1, 41, 1, 23, 43, 1, 1, 149, 29, 67, 1, 1, 1, 47, 61,
Found in Database : 5, 3, 7, 41, 23, 43, 149, 29, 67, 47, 61, 223, 163, 89, 83, 227, 367, 263, 181, 547, 1301, 229, 109,
Found in Database : 3, 5, 7, 23, 29, 41, 43, 47, 61, 67, 83, 89, 101, 103, 107, 109, 127, 149,