Proof of Riemann Conjecture Based on Contradiction between Xi-Function and Its Product Expression ()
1. Two Different Research Routes
D. Hilbert (1900) proposed 23 problems in The Second International Congress of Mathematicians and for the first time stated [1]
“…, it still remains to prove the correctness of an exceedingly important statement of Riemann, viz., that the zero points of the function
defined by the series
all have the real part 1/2, except the well-known negative integral real zeros…”
Since then it has been accepted as a classical formulation:
Riemann Hypothesis (RH). The nontrivial zeros of
all have the real part 1/2.
(But Riemann didn’t say that!) Due to Hilbert’s high prestige, the mathematicians all have focused on
. S. Smale [2] (1998) proposed 18 problems in “Mathematical problems for the next century”. RH was listed as the first. In 2000, Clay Mathematics Institute opened seven Millennium Problems, including RH, see official reviews E. Bombieri [3] (2000) and P. Sarnak [4] (2005). In the 20th century, extremely large scale computations for
confirm that RH holds (up to
) [5] [6] [7] , which have enhanced our belief. But, many scholars have different opinions about RH to be true or false, see [3] [4] . This is difficult position today.
Hilbert said [8] , “If I were to awaken after having slept for a thousand years, my first question would be: Has the Riemann hypothesis proven?” A century has passed, J. Conrey [9] (2003) pointed out that: “In my belief, RH is a genuinely arithmetic problem, likely don’t succumb to the method of analysis”. It seems that using “the hard analysis” has come to the end!
For this, we turn to the entire function
,
. But its product expression must contain complex roots, which contradicts RC. We find that the product expression is the most suitable tool for studying “no complex roots”. Finally, RC can be proved by contradiction.
In addition, RH also holds by
, but it cannot directly be proved by
. Because
is not an entire function, there is only one way, i.e. study the series summation
, which has surpassed ability of the existing analysis.
We clarify three notations used in this paper:
1) Euler
-function is analytic in the whole complex plane with a pole
.
2) Riemann took
to define
(not
used in literatures).
3) We construct
by all roots of
(Riemann had a mistake to use the same
).
2. Follow Riemann’s Thought
In Riemann’s paper, only two pages focused on RC [10] , pp. 300-302.
Euler (1737) proved product formula of primes
(1)
Taking
and
in gamma integral
and summing over n, Riemann had
where Jacobi function
satisfies
. By
, there is
The singularity
has been eliminated. Riemann got an integral representation
(2)
which has been analytically continued over the whole complex plane except for a pole
Whereas
has zeros
, called trivial zeros of
, no interest for us.
Multiplying (2) by
, Riemann directly took
and got
(3)
(Many scholars have accepted another notation
, [10] p. 17, but Riemann’s notation
is more concise in research, see (4), (5) and (6)). Inserting
into (3) and applying integration by parts twice, one has [10] p. 17,
where
. Riemann got a real function [10] pp. 301-302,
(4)
and said, “This function is finite for all finite values of t and can be developed as a power series in t2 which converges very rapidly”. Here Riemann used translation
and rotation
,
, see Figure 1, and got an even entire function
by (4). We state
Theorem A. The entire function
satisfies functional equation
Figure 1. Translation
and rotation
.
(5)
which has symmetry
and conjugate
.
Riemann continued:
“…, the function
can vanish only when the imaginary part of t lies between
and
. The number of roots of
whose real parts lie between 0 and T is about
, {remark. proved by Mangoldt, 1905}
…. One finds in fact about this many real roots within these bounds and it is very likely that all of the roots are real. One would of course like to have a rigorous proof of this, but I have put aside the research for such a proof after some fleeting vain attempts, …”
He proposed an important statement in critical strip
,
Riemann conjecture (RC). All the roots of function
are real.
Riemann finally said,
“If one denotes by
the roots of the equation
, then one can express
as
because, since the density of roots of size t grows only like
as t grows, this expression converges and for infinite t is only infinite like
; Thus it differs from
by a function of t2 which is continuous and finite for finite t and which, when divided by t2, is infinitely small for infinite t. This difference is therefore a constant, the value of which can be determined by setting
.”
We have seen that Riemann wanted to prove the following
Theorem B. Assuming that
are all roots of
, there is a product expression
(6)
It should point out that Riemann’s proof is not rigorous, but the conclusion is correct. For this, J. Hadamard [11] (1893) studied the product expression for general entire function, and (6) holds for any even entire function
of order 1, whereas
only is a special case. Why
must contain all roots? Because, let
, then
must be an even entire function of order 1 without zeros (this is very important!) and
. A simplified proof is given in [10] pp. 39-47.
3. A Mistake of Riemann and Our New Thinking
(Theorem A) functional equation
is a deep expression generated by
, which can calculate all real roots, and no complex roots are found. We have a difficult process to understand Theorem A. We once considered geometrically the peak-valley structure of
. If
is single-peak in its root interval, we for the first time have proved RC, [12] [13] [14] . How to prove the single peak? Both attempts are unsuccessful [15] [16] . It seems that the conditions are lacking. We have to consider Theorem B, so a mistake of Riemann is discovered.
(Theorem B) product expression
must contain all roots of
, which puts us in a dilemma: if there are no complex roots, then RC is assumed. If there are complex roots, it conflicts with RC. We realize that the former cannot be allowed, while the latter still has the possibility of further research, that is, to give a disproof. Theorems A and B are two different concepts, but Riemann had used the same notation
, this is a hidden mistake, which is the reason why he failed. (Unfortunately, nobody found this mistake and continued to use it as the same function). Therefore Theorem B only can be used in contradiction.
E. Bombieri [3] pointed out that “we do not have algebraic and geometric models to guide our thinking, and entirely new ideas may be needed to study these intriguing objects”. This is a valuable advice. Our research shows that geometric analysis based on Theorem A is unsuccessful, but RC can be proved by algebraic structure based on theorems A and B.
Finally we recall that if
has a finite number of complex roots, RC is proved by contradiction [17] . This paper suggests a general framework, which still leads to contradiction, even if
has infinitely many complex roots.
4. What Contradiction Do the Complex Roots Bring?
The following research can be completed on the symmetry line
(i.e.
).
Denote the complex roots
of
, where
,
(As no complex roots in
), by Theorem A,
has four conjugate complex roots
. By Theorem B,
must contain fourth degree factor
, we have on
(7)
Denote all real roots
and (true)complex roots
of
, by theorem B, we have
(8)
where
only depends on real roots
and
only depends on complex roots
, which are independent of each other. The product expression forms a multiplicative group, satisfying the exchange law and association law. This algebraic model is the base for proving RC.
We recall
, whether
has complex roots or not, the expressions (2) and (4) do not change. This is an important fact.
So our attention focused on the product expression
. There are two cases.
1). Assuming no complex roots,
,
, then
(9)
where
and
are independent of complex roots.
2). Assuming there are complex roots,
, we have
(10)
So the contradiction must appear in these t, which make
and
.
5. Properties of the Infinite Point-Set
Lemma 1. For any
, the function
is unbounded.
See [10] , p. 184. Its roots are very irregular, the first root is
, other roots
are increasingly dense, and its average spacing
is getting smaller and smaller. We have
Lemma 2. For any fixed small number
, define an infinite point-set
(11)
Its average spacing
is getting smaller and smaller.
Figure 2 shows the curves
and
in an interval
.
Lemma 3. Whether
has complex roots or not, we always have
(12)
Figure 2. Point-set
in
.
Proof. Firstly assume no complex roots, by
and lemma 2, then (12) holds. Once (12) is obtained, we found that
and
are independent of complex roots, so (12) still holds, even if
has the complex roots. The lemma is proved.
Remark 1. This reason is similar to the following fact. Under the starting condition
, Riemann had analytically continued
to (2) in the whole complex plane, except for a pole
. So this prerequisite
naturally does not work.
Lemma 4. If
has the complex roots, we have an important equality
(13)
Proof. In this case, we have
. So (13) is derived from Lemma 3.
Below we can find
to make
, which leads the contradiction.
6. Proof of Riemann Conjecture
By contradiction. If these modules
, where
is non-decreasing sequence and the convergence of
should be guaranteed. Using a transform
, we have
(14)
In
, all
and
. No matter how the complex roots are distributed, we only need to consider the first module
and its factor
(15)
Take
and
, we get
(16)
(should discuss
for multiple root
) Finally return to
, its length
is very large, in which there surely are many points
, and the contradiction is derived from Lemma 4. Therefore RC is proved.
Remark 2. As RC holds, then the positivity
, proved by Hinkkanen [18] (1997) and Lagarias [19] (1999) holds. We have
Theorem 1 [15] [16] . The strict monotone
holds for
.
This is a stronger conclusion than RC. Ancient Greek Aristotle thought, “Order and symmetry were important elements of beauty”. We say, the symmetry and order of
are mathematical beauty of Riemann conjecture.
7. Innovations and Contributions of Our Work
We look back why studying RC is so difficult. There were several mistakes. Studying RC can be compared to climbing Mount Qomolangma. The latter has two paths, i.e., along the north slope and the south slope. There are also two paths to studying RC.
The first path. Riemann (1859) had taken three steps: analytically continuation
, entire function
and product expression
. But he made a mistake in step 3. He said, “I have put aside the research for such a proof after some fleeting vain attempts”. Edwards mentioned [10] p. 164, “Siegel states quite positively that the Riemann papers contain no steps toward a proof of the Riemann hypothesis.” Now it is impossible to know what attempts are done by Riemann. Actually, Riemann had already approached to a proof of RC.
The second path. Hilbert (1900) suggested to study
, which only is the first step of Riemann. Which continuation is this? It’s also not clear. Riemann proved
and
remains unchanged when s is replaced by
(This integral is applied to derive R-S formula by him [20] ). Riemann said, “This property of the function motivated me to consider the integral
instead of the integral
in the general term of
, which leads to a very convenient expression of the function
”. Then he had gotten (2) and (4). Perhaps, Riemann thought that RH cannot be proved by
. A. Selberg [8] pointed out, “There have probably been very few attempts at proving the Riemann conjecture, because, simply, no one had ever had any really good idea for how to go about it”. We feel that studying
is a misguiding.
In order to prove RC, we have made the following innovations and contributions.
1) We have adopted entirely new method of research, Liuhui methodology (A.D.263): “computation can detect unknown” is a correct and reliable intuition, see [17] .
2) Following Riemann’s thought, we study
, but not
. The entire research can be completed on the symmetric line
(or
).
3) Find a mistake of Riemann, he used the same notation
in theorems A and B. Therefore, Theorem B can only be used by contradiction.
4) We propose a general proof by contradiction, which consists of three steps:
a) Using all real roots and (true) complex roots of
to construct
, this multiplicative group is the most suitable tool for studying “no complex roots”.
b) Define the infinite point-set
for small
. There always is
on
. If
has complex roots, then we have an important equality
.
c) No matter how the complex roots are distributed, in a large interval
we can find many points
to make
. This contradiction proves RC.
5) By
, RH is also true, but it cannot be directly proved with
, which is not an entire function. There is an essential difference between using summation or multiplication.
6) Our work opens up a broad perspective, i.e., we can propose
General RC. For a broad class of even or odd entire function
, all roots are real.
For example, Sarnak [4] discussed
and Grand-RH (related to Goldbach conjecture), which has been continued to an even entire function. Our method is useful.
Acknowledgements
The author express sincere gratitude to the referee’s for their valuable and constructive comments. Special thanks to Prof. Zhengtin Hou and Prof. Xinwen Jiang for their precious opinion in many discussions.