Download Solution to One of Landau`s Problems and Infinitely Many Prime

SOLUTION TO ONE OF LANDAU’S PROBLEMS AND
INFINITELY MANY PRIME NUMBERS OF THE FORM
ap ± b
GERMÁN ANDRÉS PAZ
Abstract. In this paper it is proved that for every positive integer
k there are infinitely many prime numbers of the form n2 + k, which
means that there are infinitely many prime numbers of the form n2 + 1.
In addition to this, in this document it is proved that if a and b are two
positive integers which are coprime and also have different parity, then
there are infinitely many prime numbers of the form ap + b, where p is a
prime number. Moreover, it is also proved that there are infinitely many
prime numbers of the form ap − b. In other words, it is proved that the
progressions ap+b and ap−b generate infinitely many prime numbers. In
particular, all this implies that there are infinitely many prime numbers
of the form 2p + 1 (since the numbers 2 and 1 are coprime and have
different parity), which means that there are infinitely many Sophie
Germain Prime Numbers. This paper also proposes an important new
conjecture about prime numbers called Conjecture C. If this conjecture
is true, then Legendre’s Conjecture, Brocard’s Conjecture and Andrica’s
Conjecture are all true, and also some other important results will be
true.
1. Introduction
FOREWORD
When I wrote my original document Números Primos de Sophie
Germain, Demostración de su Infinitud (There Are Infinitely Many
Sophie Germain Prime Numbers), I used the ‘Breusch Interval’ in order
to prove theorems and to prove that if i is any positive odd integer, then
there are infinitely many prime numbers of the form 2p + i (where p is a
prime number) and infinitely many prime numbers of the form 2p − i.
I took the Breusch Interval from the following websites created by Carlos
Giraldo Ospina (Lic. Matemáticas, USC, Cali, Colombia):
• http://www.matematicainsolita.8m.com
• http://matematicainsolita.8m.com/Archivos.htm
• http://numerosprimos.8m.com/Documentos.htm
2010 Mathematics Subject Classification. Primary 11-XX.
Key words and phrases. Landau, Landau’s Problems, near-square prime, prime, prime
number, primes in non-arithmetic progressions, Sophie Germain.
1
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GERMÁN ANDRÉS PAZ
C. G. Ospina utilized the Breusch Interval in some of his documents to
prove some specific conjectures. This author makes reference to the article
Números Primos, Balance de Nuestra Ignorancia written a long time ago
by Joaquı́n Navarro, where this person mentions the Breusch Interval and
that it was proved that there is always a prime number in this interval.
One of the people who read my document told me that they had never
heard about a so-called ‘Breusch Interval’ before, and that they were unable
to find any information on this interval in any Mathematics book.
For this reason, I wrote another document where I proved that for every
positive integer n there is always a prime number in the Breusch Interval.
To do this, I used another interval which I call ‘Iwaniec-Pintz Interval’ .
Rather than rewriting my document Números Primos de Sophie Germain,
Demostración de su Infinitud using the Iwaniec-Pintz Interval, it was easier
for me to prove that there is always a prime number in the Breusch Interval
for every positive integer n.
The present document is much more general, because here it is proved
that there are infinitely many prime numbers of the forms n2 + k, ap + b
and ap − b. In order to prove this, the Breusch Interval is used. This is
the reason why this document starts with giving a proof that the Breusch
Interval contains at least one prime number for every positive integer n.
2. Breusch Interval
i
h
9(n+3)
, where n is a positive integer, is
Definition 1. The interval n, 8
what we call ‘Breusch Interval’ .
We are going to prove that for every positive integer n there is always a
prime number in the interval mentioned above. Probably, this was already
proved a long time ago, but in this document we are going to give a clear
proof that this is true.
In order to achieve the goal, we are going to use another interval, which
we will call ‘Iwaniec-Pintz Interval’ .
3. Iwaniec-Pintz Interval
Definition 2. The Iwaniec-Pintz Interval is the interval n − nθ , n ,
where θ = 23/42.
Iwaniec
h and 23Pintz
i proved that there is always a prime number in the
42
interval n − n , n . The proof that this is true was given in their document
Primes in Short Intervals [1].
In the Iwaniec-Pintz Interval, n must be considered
asi a positive integer
h
23
42
greater than 1. This is because in the interval 1 − 1 , 1 there is no prime
number.
h
i
23
To sum up, there is always a prime number in the interval n − n 42 , n
for every positive integer n > 1.
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
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4. Modified Iwaniec-Pintz Interval
Definition 3. Working with an exponent like 23
42 may be very difficult.
Consequently,
we
define
the
‘Modified
Iwaniec-Pintz
Interval’ as the
h
i
2
interval n − n 3 , n .
We know that 23 > 23
42 , because 0.6̄ > 0.5476190... Now, if n = 1, then
23
2
23
n = n 42 = 1, but if n > 1 then n 3 > n 42 , as we will prove now:
2
3
Proof.
2
2
3
n >n
23
42
⇒
n3
n
23
42
23
>
n 42
n
23
42
⇒
5
n 42 > 1
⇒
n>
⇐
n>
⇔
n>
√
1
⇒
n>1
√
1
⇐
n>1
√
1
⇔
n>1
5
42
The reciprocal is also correct:
2
2
23
n 3 > n 42
⇐
n3
n
23
42
23
>
n 42
n
23
42
⇐
5
n 42 > 1
5
42
For this reason, it is better to say that
23
2
2
3
n >n
23
42
⇔
n3
n
23
42
>
n 42
n
23
42
⇔
5
n 42 > 1
5
42
Remark 1. In general, to prove that an inequality is correct, we can solve
that inequality step by step. If we get a result which is obviously correct,
then we can start with that correct result, ‘work backwards from there’ and
prove that the initial statement is true.
2
23
2
23
2
23
If n 3 > n 42 , then n − n 3 < n − n 42 .
Proof.
n 3 > n 42
2
⇒
23
n 3 /(−1) < n 42 /(−1)
⇒
2
⇒
2
23
−n 3 < −n 42
23
n − n 3 < n − n 42
⇒
Now, let us make a graphic of both the Iwaniec-Pintz Interval and the
Modified Iwaniec-Pintz Interval for every positive integer n > 1:
2
n − n3
|
<
23
n − n 42
<
|
{z
1.
{z
2.
• 1. Iwaniec-Pintz Interval
• 2. Modified Iwaniec-Pintz Interval
n
}
}
4
GERMÁN ANDRÉS PAZ
i
h
23
For every positive integer n greater than 1, the interval n − n 42 , n
h
i
2
contains at least one prime number. Therefore, the interval n − n 3 , n
also contains at least one prime number, according to the graphic we just
saw.
Theorem 1. If n is any
iinteger greater than 1, then the Modified
h positive
2
Iwaniec-Pintz Interval n − n 3 , n contains at least one prime number.
5. Value of n
Now, we need to know what value n needs to have so that a
Breusch Interval can contain at least one Modified Iwaniec-Pintz
Interval. We are going to use Theorem 1, which we have just
proved.
(upper endpoint of the
We do not have any way of knowing if 9(n+3)
8
Breusch Interval) is an integer or not if we
do
not
know
the value of n first.
i
h
2
The problem is that we use the interval n − n 3 , n with positive integers
only.
Let us assume y is the integer immediately preceding the number 9(n+3)
.
8
9(n+3)
In other words, y will be the largest integer that is less than 8 :
• If
9(n+3)
8
9(n+3)
8
is an integer, then
9(n+3)
8
− d = y, where d = 1.
9(n+3)
8
• If
is not an integer, then
− d = y, and in this case
we know for sure that 0 < d < 1, but we do not have any way of
knowing the exact value of d if we do not know the value of 9(n+3)
8
first.
All things considered, we have 9(n+3)
− 1 ≤ y and y <
8
that
9(n + 3)
9(n + 3)
−1≤y <
8
8
9(n+3)
.
8
This means
2
It is correct to say that if y − y 3 ≥ n, then a Breusch Interval will contain
at least one Modified Iwaniec-Pintz Interval. Let us see the graphic:
n ≤
2
y − y3
|
|
9(n + 3)
−1 ≤ y <
{z 8
}
2.
{z
3.
• 2. Modified Iwaniec-Pintz Interval
• 3. Breusch Interval
9(n + 3)
8
}
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
5
2
Let us calculate the value n needs to have so that y − y 3 ≥ n:
2
y − y3 ≥ n
2
3
9(n + 3)
9 (n + 3)
9(n + 3)
≥ n (we know that y =
−d −
−d
− d)
8
8
8
2
3
9n 9
9n 9
+ ×3−d −
+ ×3−d
8
8
8
8


≥
n
2

3
  9n 27

 9n 27
−


+
−
d
+
−
d
  8

 8
8
8
| {z }
| {z }
g
≥
n
g
In order to make the process easier, the number 27
8 − d will be called g.
27
This number will always be positive, because 8 = 3.375, and d can not be
greater than 1.
2
3
9n
9n
+g −
+g
8
8
2
3
9n
9n
+g−
+g
8
8
2
3
9n
+g
−
8
2
3
9n
−
+g
8
" 2 #
3
9n
+g
/(−1)
−
8
2
3
9n
+g
8
s
2
9n
3
+g
8
2
9n
+g
8
2
9n
9n
+2×
× g + g2
8
8
≥
n
≥
n
≥
n−
≥
−
≤
h n
i
− − g /(−1)
8
≤
n
+g
8
≤
≤
≤
9n
−g
8
n
−g
8
n
+g
8
n
3
+g
8
n 3
n 2
n
+3
×g+3
g2 + g3
8
8
8
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GERMÁN ANDRÉS PAZ
81n2 9ng
+
+ g2
64
4
81n2 + 144ng + 64g 2
64
81n2 + 144ng + 64g 2
512 ×
64
8 81n2 + 144ng + 64g 2
648n2 + 1152ng + 512g 2
≤
≤
n3
3n2 g 3ng 2
+
+
+ g3
512
64
8
n3 + 24n2 g + 192ng 2 + 512g 3
512
≤
n3 + 24n2 g + 192ng 2 + 512g 3
≤
n3 + 24n2 g + 192ng 2 + 512g 3
≤
n3 + 24n2 g + 192ng 2 + 512g 3
Now we need to know what value n needs to have so that the following
inequalities are all true:
n3 ≥ 648n2
(5.1)
24n2 g ≥ 1152ng
(5.2)
192ng 2 ≥ 512g 2
(5.3)
If these three inequalities are all true at the same time, then
n3 + 24n2 g + 192ng 2 + 512g 3
≥
648n2 + 1152ng + 512g 2
Let us calculate the values of n:
1)
n3 ≥ 648n2
⇒
24n2 g ≥ 1152ng
⇒
n3
≥ 648
n2
⇒
n ≥ 648
2)
⇒
24n2 g
≥ 1152 ⇒ 24n ≥ 1152
ng
1152
n≥
⇒ n ≥ 48
24
⇒
3)
192ng 2 ≥ 512g 2
⇒
⇒
Result 1: n ≥ 648
192ng 2
≥ 512
g2
512
n≥
⇒
192
⇒
192n ≥ 512
⇒
n ≥ 2.6̄
Result 2: n ≥ 48
Result 3: n ≥ 2.6̄
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
7
If n ≥ 648, then n > 48 and n > 2.6̄. Therefore, if n ≥ 648 then the
inequalities 5.1, 5.2 and 5.3 are all true at the same time. This means that
if n ≥ 648, then
n3 + 24n2 g + 192ng 2 + 512g 3
≥
648n2 + 1152ng + 512g 2
If we start with this true statement and we ‘work backwards from here’,
2
we prove that if n ≥ 648 is an integer, then y − y 3 ≥ n. To sum up,
n
≤
2
y − y3
<
y
<
9 (n + 3)
8
∀n ∈ Z+ such that n ≥ 648
Since hn ≥ 648 iand y > n, then y > 1. This is the reason why in the
2
interval y − y 3 , y there is always at least one prime number, according to
Theorem 1.
In other words, for every positive integer n ≥ 648 the Breusch Interval will
contain at least one Modified Iwaniec-Pintz Interval, which will contain at
least one prime number. This means that for every positive integer n ≥ 648
the Breusch Interval contains at least one prime number.
We will also prove that there is at least one prime number in the Breusch
Interval for every positive integer n such that 1 ≤ n ≤ 647:
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
Example of prime number in the interval:
2
2
3
5
5
7
7
11
11
11
11
13
13
17
17
17
17
19
19
Continues on next page...
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8
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353.25
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SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
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367.875
369
370.125
371.25
372.375
373.5
374.625
375.75
376.875
378
379.125
380.25
381.375
382.5
383.625
384.75
385.875
387
388.125
389.25
390.375
391.5
392.625
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394.875
396
397.125
398.25
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400.5
401.625
402.75
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GERMÁN ANDRÉS PAZ
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414
415.125
416.25
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418.5
419.625
420.75
421.875
423
424.125
425.25
426.375
427.5
428.625
429.75
430.875
432
433.125
434.25
435.375
436.5
437.625
438.75
439.875
441
442.125
443.25
444.375
445.5
446.625
447.75
448.875
450
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
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457.875
459
460.125
461.25
462.375
463.5
464.625
465.75
466.875
468
469.125
470.25
471.375
472.5
473.625
474.75
475.875
477
478.125
479.25
480.375
481.5
482.625
483.75
484.875
486
487.125
488.25
489.375
490.5
491.625
492.75
493.875
495
496.125
497.25
17
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GERMÁN ANDRÉS PAZ
440
441
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504
505.125
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508.5
509.625
510.75
511.875
513
514.125
515.25
516.375
517.5
518.625
519.75
520.875
522
523.125
524.25
525.375
526.5
527.625
528.75
529.875
531
532.125
533.25
534.375
535.5
536.625
537.75
538.875
540
541.125
542.25
543.375
544.5
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
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556.875
558
559.125
560.25
561.375
562.5
563.625
564.75
565.875
567
568.125
569.25
570.375
571.5
572.625
573.75
574.875
576
577.125
578.25
579.375
580.5
581.625
582.75
583.875
585
586.125
587.25
588.375
589.5
590.625
591.75
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GERMÁN ANDRÉS PAZ
524
525
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531
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603
604.125
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607.5
608.625
609.75
610.875
612
613.125
614.25
615.375
616.5
617.625
618.75
619.875
621
622.125
623.25
624.375
625.5
626.625
627.75
628.875
630
631.125
632.25
633.375
634.5
635.625
636.75
637.875
639
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
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654.75
655.875
657
658.125
659.25
660.375
661.5
662.625
663.75
664.875
666
667.125
668.25
669.375
670.5
671.625
672.75
673.875
675
676.125
677.25
678.375
679.5
680.625
681.75
682.875
684
685.125
686.25
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687.375
688.5
689.625
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691.875
693
694.125
695.25
696.375
697.5
698.625
699.75
700.875
702
703.125
704.25
705.375
706.5
707.625
708.75
709.875
711
712.125
713.25
714.375
715.5
716.625
717.75
718.875
720
721.125
722.25
723.375
724.5
725.625
726.75
727.875
729
730.125
731.25
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
23
6. Breusch Interval Conclusion
It has been proved that for every positive
integer
n there is at least one
h
i
9(n+3)
prime number in the Breusch Interval n, 8
.
+
∀n ∈ Z ,
9 (n + 3)
≥1
π n,
8
7. Theorems 2, 3 and 4
We just proved that for everyh positive iinteger n there is always a prime
number in the Breusch Interval n, 9(n+3)
.
8
Remark 2. In this document, whenever we say that a number b is between
a number a and a number c, it will mean that a < b < c, which means that
b will never be equal to a or c (the same rule will be applied to intervals).
Moreover, the number n that we will use in this document will always be a
positive integer.
Let us suppose that a is a positive integer. We need to know what value
n needs to have so that there exist a prime number p and a prime number
2n
q such that na < p < 3n
2a < q < a :
if
n
3n
2n
<p<
<q<
,
a
2a
a
then n < ap <
3n
< aq < 2n
2
In order to achieve the goal, we need to know what value n needs to
have so that there exist at least one Breusch Interval (and therefore a prime
3n
number) between na and 3n
2a and at least one Breusch Interval between 2a
2n
and a .
1) The integer immediately following the number na will be called na + d. In
other words, na + d will be the smallest integer that is greater than na :
• If na is an integer, then na + d = na + 1, because in this case we have
d = 1.
• If na is not an integer, then in the expression na +d we have 0 < d < 1,
but we do not have any way of knowing the exact value of d if we
do not know the value of na first.
The reason why we work in this way is because the number n that we will
use in the Breusch Interval will always be a positive integer.
24
GERMÁN ANDRÉS PAZ
Now, let us make the calculation:
n
a
+d+3
3n
<
8
2a
n
3n
9
+d+3 <8×
a
2a
n
24n
9
+d+3 <
a
2a
12n
n
+d+3 <
9
a
a
9n
12n
+ 9d + 27 <
a
a
12n 9n
9d + 27 <
−
a
a
3n
9d + 27 <
a
9
We need to take the largest possible value of d, which is d = 1 (if
3n
9 × 1 + 27 < 3n
a , then 9d + 27 < a for all d such that 0 < d ≤ 1):
3n
a
3n
9 + 27 <
a
3n
36 <
a
3n
> 36
a
n
36
>
a
3
n
> 12
a
9 × 1 + 27 <
Result 1:
n > 12a
So, if n > 12a then there is at least one prime number p such that
< p < 3n
2a . This means that if n > 12a, then there exists a prime number
p such that
n
a
n < ap <
3n
2
(we multiply the previous numbers by a)
2) Now, we need to calculate what value n needs to have so that there is at
least one Breusch Interval (and therefore a prime number) between 3n
2a and
2n
3n
3n
.
The
integer
immediately
following
the
number
will
be
called
a
2a
2a + d.
3n
In other words, 2a + d will be the smallest integer that is greater than 3n
2a .
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
Let us take into account that d = 1 or 0 < d < 1 depending on whether
is an integer or not, respectively.
9
3n
2a
25
3n
2a
+d+3
2n
<
8
a
2n
3n
+d+3 <8×
2a
a
3n
16n
9
+d+3 <
2a
a
27n
16n
+ 9d + 27 <
2a
a
16n 27n
9d + 27 <
−
a
2a
2 × 16n − 27n
9d + 27 <
2a
32n − 27n
9d + 27 <
2a
5n
9d + 27 <
2a
9
Now, we need to take the largest possible value of d, which is d = 1 (if
5n
9 × 1 + 27 < 5n
2a , then 9d + 27 < 2a for all d such that 0 < d ≤ 1):
5n
2a
5n
9 + 27 <
2a
5n
36 <
2a
5n
> 36
2a
5n > 36 × 2a
5n > 72a
72a
n>
5
9 × 1 + 27 <
Result 2:
n > 14.4a
So, if n > 14.4a then there is at least one prime number q such that
< q < 2n
a . This means that if n > 14.4a, then there exists a prime
number q such that
3n
2a
3n
< aq < 2n
2
(we multiply the previous numbers by a)
26
GERMÁN ANDRÉS PAZ
So far we have these two results:
Result 1:
n > 12a
Result 2:
n > 14.4a
If n > 14.4a, then n > 12a. To sum up, we have a general result:
General Result:
n > 14.4a
Theorem 2. Consequently, if a is a positive integer, then for every positive
integer n such that n > 14.4a there is always a pair of prime numbers p and
q such that n < ap < 3n
2 < aq < 2n.
Theorem 3. According to Theorem 2, we can say that if n > 14.4, then
there always exist prime numbers r and s such that n < r < 3n
2 < s < 2n.
If n > 14.4a, then there is always a pair of prime numbers p and q such
that
n < ap <
3n
< aq < 2n, according to Theorem 2
2
If n > 14.4, then there is always a pair of prime numbers r and s such
that
n<r<
3n
< s < 2n, according to Theorem 3
2
If n > 14.4a and a > 1, then n > 14.4. As a consequence, we will state
the following theorem.
Theorem 4. If a is any positive integer greater than 1, then for every
positive integer n > 14.4a there are always at least four prime numbers
p, q, r and s such that n < ap < 3n
2 < aq < 2n and simultaneously
n < r < 3n
<
s
<
2n.
2
8. Infinitely Many Primes of the Forms ap + b and ap − b
Dirichlet’s Theorem states that if a and b are two positive integers
which are coprime, then the arithmetic progression an + b contains infinitely
many prime numbers. In the expression an + b the number n is considered
as a positive integer by some authors (n ≥ 1), and it is considered as a
non-negative integer by others (n ≥ 0). In this document, n is considered
as a positive integer.
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
27
In the expression an + b, the numbers a and b must be coprime, but these
coprime numbers do not need to have different parity (they can have the
same or different parity), because the number n can be even as well as odd.
In order for ap + b (where p is a prime number) to be able to generate
infinitely many prime numbers, the numbers a and b must be coprime and
they must also have different parity. This is because even if a and b are
coprime, if these numbers are both odd then ap + b can generate at most
one prime number (only when p = 2).
When we say that the numbers a and b must have different parity, we
mean that:
• If a is even, then b must be odd.
• If a is odd, then b must be even.
Remark 3. The numbers a and b must be coprime, because if they are not,
then neither ap + b nor ap − b can generate infinitely many prime numbers,
since only composite numbers will be generated or at most a finite number
of primes will be generated. The numbers a and b must also have different
parity, otherwise only composite numbers will be generated or at most only
one prime number will be generated, as explained before.
In the article PRIMOS GEMELOS, DEMOSTRACIÓN
KMELLIZA [2], Carlos Giraldo Ospina proves that if k is any
positive even integer, then there are infinitely many prime numbers p such
that p + k is also a prime number. In particular, this author proves
that there are infinitely many twin prime numbers, solving one of
the unsolved problems in Mathematics.
We are going to prove that if a and b are two positive integers which
are coprime and have different parity, then there are infinitely many prime
numbers of the form ap + b (where p is a prime number) and infinitely many
prime numbers of the form ap − b.
In order to achieve our goal, we are going to use the same method that
C. G. Ospina used to prove Polignac’s Conjecture.
Proof.
According to Theorem 4, if a > 1 is an integer, then for every positive
integer n > 14.4a there exist prime numbers p and s, such that
n < ap <
3n
< s < 2n
2
Definition 4. The numbers ap and s form what we will call ‘pair (ap,
prime) of order b’. This is because:
28
GERMÁN ANDRÉS PAZ
• We have a pair of numbers: a number ap and a prime number s.
• The number ap is followed by the prime number s. In other words,
ap < s.
• We say that ap + b = s. In other words, we say the pair (ap, prime)
is ‘of order b’ because the difference between the numbers forming
this pair is b.
Definition 5. The set made up of all positive integers z such that z > 14.4a
will be called Set S(a).
Obviously, Set S(a) changes as the value of a changes, that is to say, Set
S(a) depends on the value of a.
Examples:
• Set S(2) is the set of all positive integers z such that z > 14.4 × 2.
In other words, Set S(2) is made up of all positive integers z such
that z > 28.8.
• Set S(3) is the set of all positive integers z such that z > 14.4 × 3.
In other words, Set S(3) is made up of all positive integers z > 43.2.
1. Let us choose any positive integer a > 1 and let us suppose that in
Set S(a) there are no pairs (ap, prime) of order b < u starting from
n = u.
The numbers n and u are positive integers which belong to Set S(a), and
the number b is a positive integer such that a and b are coprime and have
different parity.
2. Between u and 2u, n = u (and therefore in the interval [u, 2u]) there is
at least one pair (ap, prime), according to Theorem 4.
3. The difference between a number ap and a prime number which are
located between u and 2u is b < u.
4. Between u and 2u, n = u (and therefore in the interval [u, 2u])
there is at least one pair (ap, prime) of order b < u, according to
statements 2. and 3.
5. In Set S(a), starting from n = u there is at least one pair (ap, prime) of
order b < u, according to statement 4.
6. Statement 5. contradicts statement 1.
7. Therefore, no kind of pair (ap, prime) of any order b can be finite,
according to statement 6.
This means that if a > 1 is an integer, then for every positive integer b
such that a and b are coprime and have different parity there are infinitely
many prime numbers of the form ap + b, where p is a prime number. As a
consequence, we can state the following theorem.
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
29
Theorem 5. If a > 1 and b are two positive integers which are coprime and
have different parity, then there are infinitely many prime numbers of the
form ap + b, where p is a prime number.
Theorem 5 proves that there are infinitely many prime numbers
of the form 2p + 1, where p is a prime number. This is because the
numbers 2 and 1 are coprime and they also have different parity.
This proves that there are infinitely many prime numbers p such
that 2p+1 is also a prime number, which proves there are infinitely
many Sophie Germain Prime Numbers. Consequently, one of the
unsolved problems in Mathematics is now solved.
Next, we are going to prove another very important theorem.
According to Theorem 4, if a > 1 is an integer, then for every positive
integer n > 14.4a there exist prime numbers r and q such that
n<r<
3n
< aq < 2n
2
Definition 6. The numbers r and aq form what we will call ‘pair (prime,
aq) of order b’, because
• We have a pair of numbers: a prime number r and a number aq.
• The prime number r is followed by the number aq. In other words,
r < aq.
• We say that r + b = aq. In other words, we say the pair (prime, aq)
is ‘of order b’ because the difference between the numbers forming
this pair is b.
Again, the set made up of all positive integers z such that z > 14.4a will
be called Set S(a) (see Definition 5.). As we said before, Set S(a) changes
as the value of a changes.
1. Let us choose any positive integer a > 1 and let us suppose that in
Set S(a) there are no pairs (prime, aq) of order b < u starting from
n = u.
The numbers n and u are positive integers which belong to Set S(a), and
the number b is a positive integer such that a and b are coprime and have
different parity.
2. Between u and 2u, n = u (and therefore in the interval [u, 2u]) there is
at least one pair (prime, aq), according to Theorem 4.
3. The difference between a prime number and a number aq which are
located between u and 2u is b < u.
4. Between u and 2u, n = u (and therefore in the interval [u, 2u]) there is
at least one pair (prime, aq) of order b < u, according to statements 2.
and 3.
30
GERMÁN ANDRÉS PAZ
5. In Set S(a), starting from n = u there is at least one pair (prime, aq) of
order b < u, according to statement 4.
6. Statement 5. contradicts statement 1.
7. Therefore, no kind of pair (prime, aq) of any order b can be finite,
according to statement 6.
This means that if a > 1 is an integer, then for every positive integer b
such that a and b are coprime and have different parity there are infinitely
many prime numbers r such that r + b = aq. This means that there are
infinitely many prime numbers r which are prime numbers of the form aq−b,
where q is, as we said before, a prime number.
In the expression aq − b, we will change letter q for letter p and we will
state Theorem 6.
Theorem 6. If a > 1 and b are two positive integers which are coprime and
have different parity, then there are infinitely many prime numbers of the
form ap − b, where p is a prime number.
If we consider the case where a = 1 and we follow the procedures described
in this section (Section 8.), we can easily prove that for every positive even
integer k there are infinitely many prime numbers p such that p + k is also a
prime number. This is basically how C. G. Ospina already proved Polignac’s
Conjecture in his paper.
In the expression p + k, the number 1 and the positive even integer k
are coprime and have different parity (p + k = 1p + k). For this reason,
Theorems 5 and 6 can also be applied to the case where a = 1.
9. Landau’s Problems
Landau’s Problems are four problems in Number Theory concerning
prime numbers:
• Goldbach’s Conjecture: This conjecture states that every
positive even integer greater than 2 can be expressed as the sum
of two (not necessarily different) prime numbers.
• Twin Prime Conjecture: Are there infinitely many prime
numbers p such that p+2 is also a prime number? This problem was
solved by Carlos Giraldo Ospina (Lic. Matemáticas, USC,
Cali, Colombia), who proved that if k is any positive even integer,
then there are infinitely many prime numbers p such that p + k is
also a prime number.
• Legendre’s Conjecture: Is there always at least one prime number
between n2 and (n + 1)2 for every positive integer n?
• Primes of the form n2 + 1: Are there infinitely many prime
numbers of the form n2 + 1 (where n is a positive integer)?
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
31
Please see the article Primos Gemelos, Demostración Kmelliza [2],
where C. G. Ospina shows the method he used to solve the Twin Prime
Conjecture.
10. Primes of the form n2 + 1
In this document we are going to prove that there are infinitely many
prime numbers of the form n2 + 1. In order to achieve our goal, we are
going to use the same method that C. G. Ospina used to prove Polignac’s
Conjecture.
11. Theorems 7, 8 and 9
Let us suppose that n is a positive integer. We need to know what
value n needs to have so that there is always a perfect square a2 such that
n < a2 < 3n
2 .
Remark 4. We say a number is a perfect square if it is √
the square of an
integer. In other words, a number x is a perfect square if x is an integer.
Perfect squares are also called square numbers.
In general, if m is any positive integer, we need to know what value n needs
to have so that there is always a positive integer a such that n < am < 3n
2 .
We have n < am < 3n
.
This
means
that
2
3n
n < am and am <
r2
√
m 3n
m
n < a and a <
2
To sum up,
√
m
r
n<a<
m
3n
2
As we said before, the number a is
√ a positive integer.m
√Now, the integer
m
immediately
following
the
number
n
will
be
called
n√+ d. In other
√
words, m n + d is the smallest integer that is greater than m n:
√
√
√
• If m n is an integer, then m n + d = m n + 1, because in this case
we √
have d = 1.
√
• If m n is not an integer, then in the expression m n + d we have
0 < d < 1, but we do not have any way
√ of knowing the exact value
of d if we do not know the value of m n first.
32
GERMÁN ANDRÉS PAZ
Now, let us make the calculation:
√
n+d
m
r
<
m
3n
2
We need q
to take the largest possible
value of d, which is d = 1 (if
q
√
√
m 3n
m 3n
m
n+1<
n+d<
2 , then
2 for all d such that 0 < d ≤ 1):
m
√
m
√
m
√
m
r
n+1
<
1
<
1
<
1
<
1
<
1
<
1
1.5 − 1
m
1
1.5 − 1
3n
2
r
√
m 3n
− mn
2
r
√
m 3
n− mn
2
√
√
m
1.5n − m n
√
√
√
m
1.5 m n − m n
√
√ m
m
n
1.5 − 1
m
√
<
m
<
n
n
>
n
>
n
>
n
m
1
1.5 − 1
1m
√
m
m
1.5 − 1
1
√
m
m
1.5 − 1
√
m
This means that if m is a positive integer, then for every positive integer
n > m√ 1 m there is at least one positive integer a such that n < am < 3n
2 .
( 1.5−1)
Now, if n > m√ 14.4 m then n > m√ 1 m .
( 1.5−1)
( 1.5−1)
Theorem 7. Consequently, if m is a positive integer, then for every positive
integer n > m√ 14.4 m there is at least one positive integer a such that
( 1.5−1)
3n
m
n<a < 2.
Now we are going to prove that if m is any positive integer, then
> 1.
√ 1
m
( m 1.5−1)
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
33
Proof.
√
m
1
m
1.5 − 1
>
1
>
1
>
1
>
1
>
1+1
>
2
>
m
√
m
1.5 − 1
1
√
m
m
1.5 − 1
√
m
1.5 − 1
√
m
1.5 − 1
√
m
1.5
√
m
1.5
m
>
1.5
√
m
2
1
It is very easy to verify that 2m > 1.5 for every positive integer m.
Consequently, if m is any positive integer, then m√ 1 m > 1.
( 1.5−1)
√ 1
√ 14.4 m
then
> 14.4.
m > 1,
( m 1.5−1)
( m 1.5−1)
√ 14.4 m > 14.4 for every positive integer m.
( m 1.5−1)
If
This means that
Theorem 8. As a consequence, if m is any positive integer and
n > m√ 14.4 m , then n > 14.4.
( 1.5−1)
In Section 7. it was proved that for every positive integer n > 14.4 there
exist prime numbers r and s such that n < r < 3n
2 < s < 2n. This true
statement is called Theorem 3.
Theorem 9. According to Theorems 3, 7 and 8, if m is a positive integer,
then for every positive integer n > m√ 14.4 m there exist a positive integer
( 1.5−1)
a and a prime number s such that n < am < 3n
2 < s < 2n.
12. Theorems 10, 11 and 12
Now, if m is a positive integer, let us calculate what value n needs to have
m
so that there is always a positive integer a such that 3n
2 < a < 2n:
We have
3n
< am < 2n
2
This means that
34
GERMÁN ANDRÉS PAZ
3n
< am and am < 2n
2
r
√
m
m 3n
< a and a < 2n
2
To sum up,
r
m
√
3n
m
< a < 2n
2
The number a is q
a positive integer. qNow, the integer immediately
q
m 3n
m 3n
m 3n
following the number
will
be
called
+d.
In
other
words,
2
2
2 +d
q
is the smallest integer that is greater than m 3n
2 :
q
q
q
m 3n
m 3n
• If m 3n
is
an
integer,
then
+
d
=
2
2
2 + 1, because in this
caseqwe have d = 1.
q
m 3n
• If m 3n
is
not
an
integer,
then
in
the
expression
2
2 + d we have
0 < d < 1, but we do not have any way
q of knowing the exact value
of d if we do not know the value of
m
3n
2
first.
Let us make the calculation:
r
m
3n
+d
2
<
√
m
2n
take the q
largest possible value of d, which is d = 1 (if
qWe need to
√
√
m
m
3n
m 3n
2n, then
2n for all d such that 0 < d ≤ 1):
2 +1<
2 +d<
m
r
m
3n
+1
2
<
1
<
1
<
1
<
1
<
1
<
√
m
√
2n
r
3n
2
r
√
m
m 3
2n −
n
2
√
√
m
m
2n − 1.5n
√ m
√
√
√
m
m
2 n − 1.5 m n
√
√ √ m
m
m
n
2 − 1.5
m
2n −
m
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
1
√
2 − m 1.5
m
1
√
√
m
2 − m 1.5
1m
√
√ m
m
2 − m 1.5
1
√
√ m
m
2 − m 1.5
√
m
n
<
√
m
<
n
<
n
<
n
35
n
1
√
√ m
2 − m 1.5
>
m
This means that if m is a positive integer, then for every positive
integer n > m√ 1m√ m there is at least one positive integer a such that
( 2− 1.5)
1√
3n
m < 2n. Now, if n >
√
√ 14.4
√
<
a
m then n >
m.
2
( m 2− m 1.5)
( m 2− m 1.5)
Theorem 10. Consequently, if m is a positive integer, then for every
√
positive integer n > m√ 14.4
m there is at least one positive integer a such
( 2− m 1.5)
m
that 3n
2 < a < 2n.
Now we are going to prove that if m is any positive integer, then
1√
m > 1.
( 2− m 1.5)
√
m
Proof.
√
m
2−
1
√
m
m
1.5
>
1
>
1
>
1
>
1
>
1.5
>
√
m
1+
√
m
√
m
√
m
1
√
√ m
m
m
1
2 − 1.5
√
√ m
m
m
2 − 1.5
√
√
m
m
2 − 1.5
√
√
m
m
2 − 1.5
√
m
2
1.5 > 1 because 1.5 > 1m , that is to say, 1.5 > 1.
2 ≤ 2 because 2 ≤ 2m .
This means that
1+
√
m
1.5
>
2
≥
√
m
2
36
GERMÁN ANDRÉS PAZ
which proves that
1+
√
m
1.5
>
√
m
2
Therefore, if m is any positive integer, then
√
1√
1√
√
√
> 1, then m√ 14.4
> 14.4.
m
m
( m 2− m 1.5)
( 2− m 1.5)
√ 14.4
√
m > 14.4 for every positive integer m.
( m 2− m 1.5)
If
m
( m 2− m 1.5)
> 1.
This means that
Theorem 11. As a consequence, if m is any positive integer and
√
n > m√ 14.4
m , then n > 14.4.
( 2− m 1.5)
Theorem 12. According to Theorems 3, 10 and 11, if m is a positive
√
integer, then for every positive integer n > m√ 14.4
m there exist a prime
( 2− m 1.5)
m
number r and a positive integer a such that n < r < 3n
2 < a < 2n.
13. Infinitely many prime numbers of the form n2 + 1
According to Theorem 9, for every positive integer n >
√
14.4
2
( 1.5−1)
there exist a positive integer a and a prime number s such that
n < a2 < 3n
2 < s < 2n.
Definition 7. The numbers a2 and s form what we will call ‘pair
(perfect square, prime) of order k’. This is because:
• We have a pair of numbers: a perfect square a2 and a prime number
s.
• The perfect square a2 is followed by the prime number s. In other
words, a2 < s.
• We say that a2 + k = s. In other words, we say the pair
(perfect square, prime) is ‘of order k’ because the difference between
the numbers forming this pair is k.
Now we need to define some other concepts:
Definition 8. The set made up of all positive integers z such that
z > m√ 14.4 m will be called Set A(m). Examples:
( 1.5−1)
14.4
√
2.
( 1.5−1)
In other words, Set A(2) is made up of all positive integers z such
that z > 285.09...
• Set A(2) is the set of all positive integers z such that z >
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
37
14.4
√
3.
( 3 1.5−1)
In other words, Set A(3) is made up of all positive integers
z > 4751.47...
• Set A(3) is the set of all positive integers z such that z >
Definition 9. The set made up of all positive integers z such that
√
z > m√ 14.4
m will be called Set B(m).
( 2− m 1.5)
Let us prove that there are infinitely many prime numbers of the form
n2 + 1. In order to achieve the goal, we are going to use the same method
that C. G. Ospina used in his article Primos Gemelos, Demostración
Kmelliza [2] to prove Polignac’s Conjecture.
1. Let us suppose that in Set A(2) there are no pairs (perfect square, prime)
of order k < u starting from n = u.
The numbers n and u are positive integers which belong to Set A(2).
2. Between u and 2u, n = u, there is at least one pair (perfect square,
prime), according to Theorem 9.
3. The difference between two integers located between u and 2u is k < u.
4. Between u and 2u, n = u, there is at least one pair (perfect square, prime)
of order k < u, according to statements 2. and 3.
5. In Set A(2), starting from n = u there is at least one pair (perfect square,
prime) of order k < u, according to statement 4.
6. Statement 5. contradicts statement 1.
7. Therefore, no kind of pair (perfect square, prime) of any order k can be
finite, according to statement 6.
Remark 5. It is already known that for every positive integer k the
polynomial n2 + k is irreducible over R and thus irreducible over Z, since
every second-degree polynomial whose discriminant is a negative number is
irreducible over R.
According to statement 7., for every positive integer k there
are infinitely many pairs (perfect square, prime) of order k. This
means that pairs (perfect square, prime) of order 1 can not be
finite. In other words, prime numbers of the form n2 + 1 can not
be finite.
Theorem 13. All this proves that for every positive integer k there are
infinitely many prime numbers of the form n2 + k.
In general, if we use Set A(m), Set B(m) and Theorems 9 and 12 and
we use the same method, we could prove that there are infinitely many prime
numbers of the form nm + k and infinitely many prime numbers of the form
38
GERMÁN ANDRÉS PAZ
nm − k for certain values of m and k (we only have to take into account the
cases where the polynomials nm + k and nm − k are irreducible over Z).
14. Final Conclusion
We will restate the three most important theorems that were proved in
this document (see Sections 8. and 13.):
Theorem 5: If a and b are two positive integers
which are coprime and have different parity, then
there are infinitely many prime numbers of the
form ap + b, where p is a prime number.
Theorem 6: If a and b are two positive integers
which are coprime and have different parity, then
there are infinitely many prime numbers of the
form ap − b, where p is a prime number.
Theorem 13: For every positive integer k there are
infinitely many prime numbers of the form n2 + k,
which means there are infinitely many primes of
the form n2 + 1.
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
39
Appendix A. New conjecture
Conjecture 1. If n is any positive integer and we take n consecutive
integers located between n2 and (n + 1)2 , then among those n integers there
is at least one prime number. In other words, if a1 , a2 , a3 , a4 , . . . , an are n
consecutive integers such that n2 < a1 < a2 < a3 < a4 < ... < an < (n + 1)2 ,
then at least one of those n integers is a prime number. This conjecture will
be called Conjecture C.
• Legendre’s Conjecture
It is very easy to verify that the amount of integers located between n2
and (n + 1)2 is equal to 2n.
Proof.
(n + 1)2 − n2 = 2n + 1
n2 + 2n + 1 − n2 = 2n + 1
2n + 1 = 2n + 1
We need to exclude the number (n + 1)2 because we are taking into
consideration the integers that are greater than n2 and smaller than (n + 1)2 :
2n + 1 − 1 = 2n
According to this, between n2 and (n + 1)2 there are two groups of n
consecutive integers each that do not have any integer in common. Example
for n = 3:
(3)2
10
|
11
{z
13
|
12}
Group A
(n consecutive integers)
|
14
{z
(3 + 1)2
15}
Group B
(n consecutive integers)
{z
2n consecutive integers
}
Group A and Group B do not have any integer in common. According
to Conjecture C, Group A contains at least one prime number and
Group B also contains at least one prime number, which means that between
32 and (3 + 1)2 there are at least two prime numbers. This is true because
the numbers 11 and 13 are both prime numbers.
All this means that if Conjecture C is true, then there are at least two
prime numbers between n2 and (n + 1)2 for every positive integer n. As a
result, if Conjecture C is true, then Legendre’s Conjecture is also
true.
40
GERMÁN ANDRÉS PAZ
• Brocard’s Conjecture
This conjecture states that if pn and pn+1 are consecutive prime numbers
greater than 2, then between (pn )2 and (pn+1 )2 there are at least four prime
numbers.
Since 2 < pn < pn+1 , we have pn+1 − pn ≥ 2. This means that there is at
least one positive integer a such that pn < a < pn+1 . As a result, there is
at least one positive integer a such that (pn )2 < a2 < (pn+1 )2 .
Conjecture C states that between (pn )2 and a2 there are at least two
prime numbers and that between a2 and (pn+1 )2 there are also at least two
prime numbers. In other words, if Conjecture C is true then there are at
least four prime numbers between (pn )2 and (pn+1 )2 . As a consequence,
if Conjecture C is true then Brocard’s Conjecture is also true.
• Andrica’s Conjecture
√
√
This conjecture states that pn+1 − pn < 1 for every pair of consecutive
prime numbers pn and pn+1 (of course pn < pn+1 ).
Obviously, every prime number is located between two consecutive perfect
squares. If we take any prime number pn , which is obviously located between
n2 and (n + 1)2 , two things may happen:
Case 1. The number pn is located among the first n consecutive integers
that are located between n2 and (n + 1)2 . These n integers form what we
call Group A, and the following n integers form what we call Group B, as
shown in the following graphic:
n2
<
•
|
•
...
{z
•
•}
•|
Group A
(n consecutive integers)
|
•
...
{z
•
•}
<
(n + 1)2
Group B
(n consecutive integers)
{z
2n consecutive integers
}
If pn is located in Group A and Conjecture C is true, then pn+1 is either
√
√
located in Group A or in Group B. In both cases we have pn+1 − pn < 1,
q
√
√
√
because (n + 1)2 − n2 = 1 and the numbers pn+1 and pn are closer
q
√
to each other than (n + 1)2 in relation to n2 .
Case 2. The prime number pn is located in Group B.
If pn is located in Group B and Conjecture C is true, it may happen that
pn+1 is also located in Group B. In this case it is very easy to verify that
√
√
pn+1 − pn < 1, as explained before.
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
41
Otherwise, if pn+1 is not located in Group B, then pn+1 is
located in Group C. In this case the largest value pn+1 can have
is pn+1 = (n + 1)2 + n + 1 = n2 + 2n + 1 + n + 1 = n2 + 3n + 2 and
the smallest value pn can have is pn = n2 + n + 1 (in order to make the
process easier, we are not taking into account that in this case the numbers
pn and pn+1 have different parity, so they can not be both prime at the same
time).
√
√
This means that the largest possible difference between pn+1 and pn
√
√
√
√
is pn+1 − pn = n2 + 3n + 2 − n2 + n + 1. Let us look at the following
graphic:
maximum distance between pn and pn+1
2
n <
...
4
|
•
...
{z
•
Group B
(n consecutive integers)
• < (n + 1)2 < •|
}
•
... {z •
•
Group C
(n+1 consecutive integers)
2}
4 = n2 + n + 1 = pn
2 = n2 + 3n + 2 = pn+1
√
√
It is easy to prove that n2 + 3n + 2 − n2 + n + 1 < 1.
Proof.
p
p
n2 + 3n + 2 − n2 + n + 1
p
n2 + 3n + 2
<
1
<
n2 + 3n + 2
<
n2 + 3n + 2
<
n2 + 3n + 2 − n2 − n − 1
<
2n + 1
<
2n
<
n
<
n
<
p
1 + n2 + n + 1
2
p
1 + n2 + n + 1
p
1 + 2 n2 + n + 1 + n2 + n + 1
p
1 + 2 n2 + n + 1
p
1 + 2 n2 + n + 1
p
2 n2 + n + 1
√
2 n2 + n + 1
2
p
2
n +n+1
n2
<
n2 + n + 1
which is true for every positive integer n.
We can see that even when the difference between pn+1 and pn is the
√
√
largest possible difference, we have pn+1 − pn < 1. If the difference
between pn+1 and pn were smaller, then of course it would also happen that
√
√
pn+1 − pn < 1.
42
GERMÁN ANDRÉS PAZ
According to Cases 1. and 2., if Conjecture C is true then
Andrica’s Conjecture is also true.
To conclude, if Conjecture C is true, then Legendre’s
Conjecture, Brocard’s Conjecture and Andrica’s Conjecture are
all true.
• Possible new interval
2It is easy to2 verify that
if Conjecture C is true, then in the interval
n + n + 1, n + 3n + 2 (see the graphics shown before) there are at least
two prime numbers for every positive integer n.
The number n2 + n + 1 will always be an odd integer.
Proof.
• If n is even, then n2 is also even. Then we have
(even integer + even integer) + 1 = even integer + odd integer =
= odd integer
• If n is odd, then n2 is also odd. Then we have
(odd integer + odd integer) + 1 = even integer + odd integer =
= odd integer
Since the number n2 + n + 1 will always be an odd integer, then it may
be prime or not.
Now, the number n2 + 3n + 2 can never be prime because this number
will always be an even integer (and it will be greater than 2).
Proof.
• If n = 1 (smallest value n can have), then n2 + 3n + 2 = 1 + 3 + 2 = 6.
• If n is even, then n2 and 3n are both even integers. The number 2 is also
an even integer, and we know that
even integer + even integer + even integer = even integer
• If n is odd, then n2 and 3n are both odd integers, and we know that
SOLUTION TO ONE OF LANDAU’S PROBLEMS AND INFINITELY MANY...
43
(odd integer + odd integer) + even integer =
= even integer + even integer = even integer
From all this we deduce that if Conjecture C is true, then the maximum
distance between two consecutive prime numbers is the one from the number
n2 + n + 1 to the number n2 + 3n + 2 − 1 = n2 + 3n + 1, which means
that in the interval [n2 + n + 1, n2 + 3n + 1] there are at least two prime
numbers. In other words, in the interval [n2 + n + 1, n2 + 3n] there is at
least one prime number.
The difference between the numbers n2 + n + 1 and n2 + 3n
2
2
is n2 + 3nj− (n2 + n + 1)
k = n + 3n − n − n − 1 = 2n − 1. In addition
√
n2 + n + 1
= n.
This means that in the interval
to this,
h
j√
k
i
n2 + n + 1, n2 + n + 1 + 2
n2 + n + 1 − 1 there is at least one prime
number. √In other words, if a = n2 + n + 1 then the interval
[a, a + 2 b ac − 1] contains at least one prime number.
The symbol bc represents the floor function. The floor function of a given
number is the largest integer that is not greater than that number. For
example, bxc is the largest integer that is not greater than x.
Now, if Conjecture C is true, then the following statements are all true:
√
Statement 1. If a is a perfect square, then in the interval [a, a + b ac]
there is at least one prime number.
2
2
2
Statement 2. If a is an integer
√ such that n < a ≤ n + n + 1 < (n + 1) ,
then in the interval [a, a + b ac − 1] there is at least one prime number.
2
2
2
Statement 3. If a is an integer
√ such that n < n + n + 2 ≤ a < (n + 1) ,
then in the interval [a, a + 2 b ac − 1] there is at least one prime number.
√
√
We know that a + 2 b ac − 1 ≥ a + b ac.
Proof.
a+2
√ √ a −1≥a+
a
⇔
⇔
2
⇔
2
√ √ a −1≥
a
√ √ a ≥
a +1
√ √ √ a +
a ≥
a +1
which is true for every positive integer a.
⇔
⇔
⇔
√ a ≥ 1,
44
GERMÁN ANDRÉS PAZ
√
√
And we also know that a + 2 b ac − 1 > a + b ac − 1.
Proof.
a+2
√ √ a −1>a+
a −1
⇔
2
√ √ a >
a ,
which is obviously true for every positive integer a.
√
All this means that the interval [a, a + 2 b ac − 1] can be applied to the
number a from Statement 1., to the number a from Statement 2. and to the
number a from Statement 3.
Therefore, if n is any positive
√ integer and Conjecture C is true,
then in the interval [n, n + 2 b nc − 1] there is at least one prime
number (we change letter a for letter n). According to this,√we can
also say that if Conjecture C is true then in the interval [n, n + 2 n − 1]
there is always a prime number for every positive integer n.
Now. . . how can we prove Conjecture C?
===================0===================
References
[1]
Iwaniec, H. and Pintz, J. Primes in Short Intervals. Monatsh. f. Math. 98, 115-143,
1984.
[2]
Carlos Giraldo Ospina. Primos Gemelos, Demostración Kmelliza (in Spanish).
http://matematicainsolita2.8m.com/ar90.htm
E-mail address, Germán Andrés Paz: germanpaz [email protected]
(2000) Rosario, Santa Fe, Argentina