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Development of Algorithmic Constructions

06:14:24 20.Apr 2024

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Polynom = x^2+25x-97

1. Algorithm

a:=1;
b:=25;
c:=-97;


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+25x-97

f(0)=97
f(1)=71
f(2)=43
f(3)=13
f(4)=19
f(5)=53
f(6)=89
f(7)=127
f(8)=167
f(9)=11
f(10)=23
f(11)=1
f(12)=347
f(13)=397
f(14)=449
f(15)=503
f(16)=1
f(17)=617
f(18)=677
f(19)=739
f(20)=73
f(21)=79
f(22)=937
f(23)=1
f(24)=83
f(25)=1153
f(26)=1229
f(27)=1307
f(28)=1
f(29)=113
f(30)=1553
f(31)=149
f(32)=157
f(33)=1
f(34)=1
f(35)=2003
f(36)=2099
f(37)=1
f(38)=2297
f(39)=2399
f(40)=2503
f(41)=2609
f(42)=1
f(43)=257
f(44)=2939
f(45)=1
f(46)=3169
f(47)=173
f(48)=3407
f(49)=3529
f(50)=281
f(51)=3779
f(52)=3907
f(53)=367
f(54)=379
f(55)=331
f(56)=193
f(57)=199
f(58)=1
f(59)=1
f(60)=5003
f(61)=271
f(62)=5297
f(63)=419
f(64)=509
f(65)=523
f(66)=311
f(67)=6067
f(68)=479
f(69)=6389
f(70)=6553
f(71)=6719
f(72)=1
f(73)=7057
f(74)=7229
f(75)=673
f(76)=1
f(77)=7757
f(78)=7937
f(79)=353
f(80)=1
f(81)=653
f(82)=8677
f(83)=8867
f(84)=9059
f(85)=487
f(86)=859
f(87)=877
f(88)=229
f(89)=773
f(90)=10253
f(91)=10459
f(92)=10667
f(93)=1
f(94)=853
f(95)=1
f(96)=11519
f(97)=1
f(98)=1087
f(99)=641

b) Substitution of the polynom
The polynom f(x)=x^2+25x-97  could be written as f(y)= y^2-253.25 with x=y-12.5

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+12.5
f'(x)>2x+24

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 : 97, 71, 43, 13, 19, 53, 89, 127, 167, 11, 23, 1, 347, 397, 449, 503, 1, 617, 677, 739,
Found in Database : 97, 71, 43, 13, 19, 53, 89, 127, 167, 11, 23, 347, 397, 449, 503, 617, 677, 739, 73, 79, 937, 83, 1153, 1229, 1307, 113, 1553, 149, 157, 2003, 2099, 2297, 2399,
Found in Database : 11, 13, 19, 23, 43, 53, 71, 73, 79, 83, 89, 97, 113, 127, 149,