F1000Research F1000Research 2046-1402 F1000 Research Limited London, UK 10.12688/f1000research.172983.1 Research Article Articles New Results About Imaginary Value of the Second Axiomatic Zero of the Zeta Function and the Fifth Bernoulli Number

[version 1; peer review: awaiting peer review]

 Tuma Ahmed Adnan Conceptualization 1 Salman Shakir M. Conceptualization https://orcid.org/0000-0002-0216-6712 a 2 Mathematics, University of Baghdad College of Education for Pure Science Ibn Al-Haitham, Baghdad, Baghdad Governorate, Iraq Mathematics, Department of Mathematics, College of Basic Education, University of Diyala, Baqubah, Iraq basicmath6te@uodiyala.edu.iq

No competing interests were disclosed.

 10 1 2026 2026 15 46 23 12 2025 Copyright: © 2026 Tuma AA and Salman SM 2026 This is an open access article distributed under the terms of the Creative Commons Attribution Licence, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

This paper introduces two theorems in number theory about zeta at − 4 and the fifth Bernoulli number B 5 , showing that the zeta function ζ ( − 4 ) has two complex values and that the fifth Bernoulli number is not zero but alsohas two complex values.

The current research concludes that the zeta function ζ ( − 4 ) has two imaginary values that are different from what was previously known to mathematicians, which was believed to be equal to zero, whereas, ζ ( − 4 ) = - i 558 , ζ ( − 4 ) = - i 2232

Therefore, the fifth Bernoulli number is not equal to zero and has two values: B 5 = − 5 i 558 and B 5 = − 5 i 2232 .

 Zeta function Bernoulli numbers trivial zeros. The author(s) declared that no grants were involved in supporting this work. 1. Introduction

In 1713 Swedish mathematician Jacob Bernoulli and Japanese mathematician Seki Takakazu independently discovered numbers now called Bernoulli numbers through their work on the power series of integers 1 m + 2 m + 3 m + … + n m . After their work, they accidentally discovered the values of these numbers and, through law 1 : S m ( n − 1 ) = 1 m + 1 ∑ k = 1 m ( m + 1 k ) B k n m − k + 1 determined the final values of these numbers, and concluded that all odd Bernoulli numbers are zero except for the first Bernoulli number B 1 = − 1 2 , and even Bernoulli numbers have the following values 2 : B 0 = 1 , B 1 = − 1 2 , B 2 = 1 6 , B 3 = 0 , B 4 = − 1 30 , B 5 = 0 , B 6 = 1 42

In 1734, Swiss mathematician Leonardo Euler formulated the zeta function: ζ ( s ) = ∑ n = 1 ∞ 1 n s , where, s ∈ ℝ .

Euler defined a zeta function in the domain of real numbers. He succeeded in cleverly linking the zeta function to prime numbers through the law that he derived. ζ ( s ) = ∏ p 1 ( 1 − p − s )

He also succeeded in solving the Basel problem, which was posed in Basel City, and provided proof of the result using the zeta function, where 3 ζ ( 2 ) = 1 + 1 2 2 + 1 3 2 + 1 4 2 + … = π 2 6

By studying the zeta function and linking it with Bernoulli numbers, Euler succeeded in deducing two of the most important laws. Through this study, he succeeded in linking the Bernoulli numbers with the zeta function. ζ ( 2 k ) = ( − 1 ) k + 1 ⋅ ( 2 π ) 2 k ⋅ B 2 k 2 ( 2 k ) ! ; k = 1 , 2 , 3 , … ζ ( − k ) = ( − 1 ) k ⋅ B k + 1 k + 1 ; k = 0 , 1 , 2 , 3 , …

From this, he concluded that the zeta function at − 1 equals − 1 12 . 4 ζ ( − 1 ) = 1 + 2 + 3 + 4 + … = − 1 12

if k = 1 then, ζ ( − 1 ) = ( − 1 ) 1 ⋅ B 2 2 = − 1 / 6 2 = − 1 12

In 1859, German scientist Bernard Riemann extended the range of functions to include imaginary numbers as follows: ζ ( s ) = ∑ n = 1 ∞ 1 n s , where, s ∈ ℂ .

Riemann extended the range of the function’s inputs to include imaginary numbers and stated that all non-trivial zeros lie on the critical line, which was named after him for the results he provided on this function. The Riemann conjecture is considered to be an unsolved mathematical problem. 5

Ramanujan presented a proof in which he proved that the zeta function at the negative number − 1 is equal to − 1 12 , where ζ ( − 1 ) = 1 + 2 + 3 + 4 + … = − 1 12

Using special sums in a specific manner, as in, 6 It should also be emphasized that Euler presented a proof in which he showed that the zeta function at − 1 equals − 1 12 , based on Bernoulli numbers, and Ramanujan proved that the zeta function at − 1 equals − 1 12 using sums.

Another researcher presented a refutation of what are called axiomatic zeros, whose values depend on Bernoulli numbers 7 using partial sums. The researcher presented a proof to calculate the value of the zeta function at the number − 2 and proved that its value is not equal to zero as Euler extracted it using his second law based on Bernoulli numbers, but rather equals an imaginary value of i 84 . The researcher also proved that the third Bernoulli number is not zero based on Euler’s second law, considering that zeta is known and the Bernoulli number is unknown, thus concluding that the third Bernoulli number is equal to i 28 .

 2. The main results

In this section, two theorems are proved and show that the zeta function at ( − 4 ) has two complex values and that the fifth Bernoulli number B 5 is not zero but has two complex values − 5 i 558 and − 5 i 2232 .

Zeta function at − 4 has two values - i 558 and - i 2232

ζ ( s ) = ∑ n = 1 ∞ 1 n s , where s ∈ ℝ according to the Euler definition. ζ ( − 4 ) = ∑ n = 1 ∞ 1 n − 4 = ∑ n = 1 ∞ n 4 = 1 + 16 + 81 + 256 + 625 + 1296 + …

From, 8 η ( s ) = ∑ n = 1 ∞ ( − 1 ) n + 1 n s ; s = − 4 η ( − 4 ) = ∑ n = 1 ∞ ( − 1 ) n + 1 n − 4 = 1 − 16 + 81 − 256 + 625 − 1296 + … ζ ( − 4 ) − η ( − 4 ) = 32 + 512 + 2592 + … = 32 ( 1 + 16 + 81 + … ) ζ ( − 4 ) − 32 ζ ( − 4 ) = η ( − 4 ) ζ ( − 4 ) = − η ( − 4 ) 31

Now, η ( − 4 ) = 1 − 16 + 81 − 256 + 625 − 1296 + … 1 = [ 1 ( − 1 ) 1 ( − 1 ) ] 1 − 16 = − 15 = { [ ( 1 + 2 ) ( 4 ) + ( 1 + 2 ) ( 1 ) ] [ ( − 1 ) 1 + 1 ( − 1 ) ] } 1 − 16 + 81 = 66 = { [ ( 1 + 2 + 3 ) ( 9 ) + ( 1 + 2 + 3 ) ( 1 + 1 ) ] [ ( − 1 ) 1 + 1 + 1 ( − 1 ) ] } 1 − 16 + 81 − 256 = − 190 = { [ ( 1 + 2 + 3 + 4 ) ( 16 ) + ( 1 + 2 + 3 + 4 ) ( 1 + 1 + 1 ) ] [ ( − 1 ) 1 + 1 + 1 + 1 ( − 1 ) ] } 1 − 16 + 81 − 256 + 625 = 435 = { [ ( 1 + 2 + 3 + 4 + 5 ) ( 25 ) + ( 1 + 2 + 3 + 4 + 5 ) ( 1 + 1 + 1 + 1 ) ] [ ( − 1 ) 1 + 1 + 1 + 1 + 1 ( − 1 ) ] } and so on, η ( − 4 ) = { ( 1 + 2 + 3 + … ) [ ( 1 + 2 + 3 + … ) + ( 1 + 2 + … ) ] + [ ( 1 + 2 + 3 + … ) ( 1 + 1 + … ) ] } [ ( − 1 ) 1 + 1 + 1 + … ( − 1 ) ] η ( − 4 ) = { ζ ( − 1 ) [ ζ ( − 1 ) + ζ ( − 1 ) ] + ζ ( − 1 ) ζ ( 0 ) } [ ( − 1 ) ζ ( 0 ) ( − 1 ) ] η ( − 4 ) = { − 1 12 [ − 1 12 + − 1 12 ] + − 1 12 ∙ − 1 2 } [ ( − 1 ) − 1 2 ( − 1 ) ] η ( − 4 ) = i 18

Put in Equation (3) we get ζ ( − 4 ) = − i / 18 31 = − i 558

By Equation (4): 1 = [ 1 ( − 1 ) 1 ( − 1 ) ] 1 − 16 = − 15 = { [ ( 1 + 2 ) ( 1 + 2 + 1 + 2 − 2 ) + ( 1 + 2 ) ( 1 ) ] [ ( − 1 ) 1 + 1 ( − 1 ) ] } 1 − 16 + 81 = 66 = { [ ( 1 + 2 + 3 ) ( 1 + 2 + 3 + 1 + 2 + 3 − 3 ) + ( 1 + 2 + 3 ) ( 1 + 1 ) ] [ ( − 1 ) 1 + 1 + 1 ( − 1 ) ] } 1 − 16 + 81 − 256 = − 190 = { [ ( 1 + 2 + 3 + 4 ) ( 1 + 2 + 3 + 4 + 1 + 2 + 3 + 4 − 4 ) + ( 1 + 2 + 3 + 4 ) ( 1 + 1 + 1 ) ] [ ( − 1 ) 1 + 1 + 1 + 1 ( − 1 ) ] } 1 − 16 + 81 − 256 + 625 = 435 = { [ ( 1 + 2 + 3 + 4 + 5 ) ( 1 + 2 + 3 + 4 + 5 + 1 + 2 + 3 + 4 + 5 − 5 ) + ( 1 + 2 + 3 + 4 + 5 ) ( 1 + 1 + 1 + 1 ) ] [ ( − 1 ) 1 + 1 + 1 + 1 + 1 ( − 1 ) ] } and so on, η ( − 4 ) = { ( 1 + 2 + 3 + … ) [ ( 1 + 2 + 3 + … ) + [ 1 + 2 + 3 + … ) − ( 1 + 1 + 1 + … ) ] ] + [ ( 1 + 2 + 3 + … ) ( 1 + 1 + … ) ] } [ ( − 1 ) 1 + 1 + 1 + … ( − 1 ) ] η ( − 4 ) = { ( ζ ( − 1 ) ) [ ( ζ ( − 1 ) ) + [ ζ ( − 1 ) ) − ( ζ ( 0 ) ) ] ] + [ ( ζ ( − 1 ) ) ( ζ ( 0 ) ) ] } [ ( − 1 ) ζ ( 0 ) ( − 1 ) ] η ( − 4 ) = { ( − 1 12 ) [ ( − 1 12 ) + [ − 1 12 − ( − 1 2 ) ] ] + [ ( − 1 12 ) ( − 1 2 ) ] } [ ( − 1 ) − 1 2 ( − 1 ) ] η ( − 4 ) = i 72

Put in Equation (3) we get ζ ( − 4 ) = − i / 72 31 = − i 2232

The fifth Bernoulli number, B 5 is not zero but has two values: − 5 i 558 and − 5 i 2232 .

Using (2) ζ ( − k ) = ( − 1 ) k B k + 1 k + 1 being ζ ( − 4 ) = ( − 1 ) 4 B 5 5

Since ζ ( − 4 ) = − i 558 from ( 5 ) , we get, B 5 = − 5 i 558 .

And, Since ζ ( − 4 ) = − i 2232 from ( 7 ) , we get, B 5 = − 5 i 2232

3. Conclusions

We conclude from the current research that the zeta function ζ ( − 4 ) has two imaginary values that are different from what was previously known to mathematicians, which was believed to be equal to zero, whereas, ζ ( − 4 ) = - i 558 , ζ ( − 4 ) = - i 2232

Therefore, the fifth Bernoulli number is not equal to zero and has two values: B 5 = − 5 i 558 and B 5 = − 5 i 2232 .

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