Riemann Hypothesis
Recommended article: 【Modern Mathematics】 Table of Contents of Modern Mathematics
1. Overview
1. Overview
⑴ Riemann zeta function
① Definition: $\displaystyle \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$
② The Riemann zeta function can be expressed as an Euler product.
○ The Riemann zeta function above converges for $\operatorname{Re}(s)>1$ and has a pole at $s=1$.
○ The Riemann hypothesis asks whether the real part of every nontrivial zero is $1/2$.
③ A main identity
○ From the above equation, we obtain $\displaystyle \lim_{s\to 1}(s-1)\zeta(s)=1$.
⑵ Von Mangoldt function $\Lambda(n)$
⑶ Dirichlet $L$-function
① $\chi$: A Dirichlet character used to distinguish residue classes modulo a given integer $q$. For $q=3$, it is given as follows.
② Combining it with the von Mangoldt function yields the following equation.
⑷ Hecke $L$-function: A generalization to Dirichlet $L$-function
⑸ Möbius function $\mu(n)$
⑹ Mertens function $M(x)$
2. Prime Number Theorem
⑴ Statement: A theorem concerning the density of prime numbers.
① For example, when $x=10^6$, the actual number of primes is 78,498, whereas $x/\log x$ is approximately 72,382.
② From the asymptotic equivalence above, one might expect that a similar density could arise from other rules defining subsets of natural numbers, not necessarily just prime numbers.
⑵ Proof 1. Hadamard (1896): Proved the existence of a zero-free region for the zeta function.
⑶ Proof 2. de la Vallée Poussin (1896): Controlled the locations of zeros and estimated the distribution of prime numbers.
⑷ Proof 3. Erdős (1948–1949): Used the fact that sufficiently many prime numbers exist even in short intervals.
⑸ Proof 4. Selberg (1948–1949): Derived a symmetric formula involving prime numbers and pairs of primes.
⑹ Proof 5. Newman (1980): Derived the asymptotic distribution of prime numbers simply from analytic continuation.
⑺ Proof 7. Modern analytic proof
① Step 1. Zeta function inequality: For $\sigma>1$.
② Step 2. The zeta function has no zeros on $\operatorname{Re}(s)=1$: $\zeta(1+it)\neq 0$ (for $t\neq 0$).
○ Since we have already established $\displaystyle \lim_{s\to 1}(s-1)\zeta(s)=1$, the point $s=1$ is a pole rather than a zero: This also gives $\zeta(1+\varepsilon)\sim 1/\varepsilon$.
○ Since $\zeta(1+\varepsilon)\sim 1/\varepsilon$, while $\zeta(1+\varepsilon+it_0)$ is assumed to approach a zero, a certain tension arises, requiring $m\geq 1$.
○ Although $\zeta(1+\varepsilon+it_0)$ is associated with a zero, it is unlikely that $\zeta(1+\varepsilon+2it_0)$ has a related zero or pole, so it is treated as $O(1)$.
○ Proof by contradiction: Assuming that a zero exists leads to a contradiction with the inequality obtained in Step 1 → Therefore, no such zero exists.
○ One may also interpret this as follows: Since infinitely many different trigonometric identities can be chosen in Step 1, they impose so many constraints that the existence of a zero becomes impossible.
③ Step 3. Wiener–Ikehara theorem
○ Theorem: If $\displaystyle F(s)=\sum_{n=1}^{\infty}\frac{a_n}{n^s}$, $a_n\geq 0$, and $F(s)-\frac{A}{s-1}$ admits an analytic continuation to a neighborhood of $\operatorname{Re}(s)\geq 1$, then the following holds.
○ Substituting $a_n=\Lambda(n)$, the von Mangoldt function, and $A=1$ yields the final conclusion.
⑻ Prime Number Theorem for Arithmetic Progressions: For fixed $q$, $\gcd(a,q)=1$, and Euler’s totient function $\varphi(q)$.
3. Goldbach’s Conjecture
⑴ Strong Goldbach’s Conjecture
① The claim that every even integer greater than 2 can be expressed as the sum of two prime numbers. It has not yet been fully proven.
⑵ Weak Goldbach’s Conjecture (► Video Explanation)
① Overview
○ Statement: Every odd integer greater than 5 can be expressed as the sum of three prime numbers.
○ Proved by Harald Andrés Helfgott in 2013. (ref)
○ Example 1. $7=2+2+3$
○ Example 2. $9=3+3+3$
② Step 1. Analytic Proof for Sufficiently Large $N$
○ There exists some $N_0$ such that the statement can be proved analytically for all $N \geq N_0$, using analytic methods to estimate the distribution of prime numbers.
○ Specifically, every odd integer $N\geq 10^{27}$ can be expressed as the sum of three prime numbers.
③ 1-1. $R(N)=$ The number of ordered representations of $N$ as the sum of three prime numbers.
○ The statement to be proved is $R(N)>0$.
④ 1-2. Express the distribution of prime numbers using a Fourier series.
○ This identity is an exact mathematical equality, not an estimate or approximation.
○ Major arcs $\mathfrak{M}$: Subintervals of the integration domain $[0,1]$ for $\alpha$ that lie near rational numbers with small denominators. The distribution of prime numbers can be calculated relatively precisely in these regions.
○ Minor arcs $\mathfrak{m}$: The remaining intervals. Although accurately calculating individual values is difficult, the overall magnitude can be bounded by exploiting cancellation due to oscillation.
⑤ 1-3. Lower-bound estimate for the major arcs
○ Intuitive understanding
○ Number of solutions to $a+b+c=N$, where $a,b,c\geq 1$: $\displaystyle \binom{N-1}{2}=\frac{(N-1)(N-2)}{2}\approx\frac{N^2}{2}$
○ According to the Prime Number Theorem, $\pi(N)\sim N/\log N$. Therefore, for large numbers of order $N$, the probability that an arbitrary integer is prime is approximately $P(\text{prime})\approx 1/\log N$.
○ $P(a,b,c\text{ are all prime})\approx\frac{1}{(\log N)^3}$
○ $\therefore$ $\displaystyle R(N)\approx\frac{N^2}{2(\log N)^3}$ (not rigorous)
○ Since the primality of three integers is not completely independent, we must multiply by a singular series $\mathfrak{S}(N)$: The singular series $\mathfrak{S}(N)$ below can be intuitively derived from a combinatorial perspective by calculating the ratio of two probabilities for each prime $p$ and then multiplying these ratios over all primes.
○ $\therefore$ $\displaystyle R(N)\sim\mathfrak{S}(N)\frac{N^2}{2(\log N)^3}$ (not rigorous)
○ However, this formula is an asymptotic expression for sufficiently large odd integers $N$, and its rigorous justification is provided by the circle method.
○ For $\alpha\approx a/q$, the integers $a$ and $q$ are coprime, and $q$ is a positive integer (with $q$ relatively small).
○ Prime Number Theorem for Arithmetic Progressions: $\displaystyle \pi(N;q,a)\sim\frac{N}{\varphi(q)\log N}$
○ In other words, for $q=3,5,7,101$, the numbers of admissible residue classes $\varphi(q)$ are $2,4,6,100$, respectively, and the proportion of primes distributed among each residue class is the reciprocal of that number.
○ When $q$ becomes large, two problems arise.
○ Problem 1. Since the number of primes in each group decreases, stronger results are required to estimate the distribution of each group with the same relative error.
○ Problem 2. The complex terms in the exponential sum become distributed in more diverse directions and may cancel one another.
○ By exploiting the regular phase structure associated with small denominators $q$, one can analytically establish an upper bound on the absolute value of the error.
○ Conclusion: $\displaystyle I_{\mathrm{major}}(N)\geq\text{positive main term}-\sup\left|E(N)\right|$
⑥ 1-4. Upper-bound estimate for the minor arcs
○ Parseval’s identity in Fourier analysis
○ Triangle inequality for integrals
○ Intuitively, on the minor arcs, complex exponential terms rotate in different directions and exhibit substantial cancellation, making the maximum magnitude of $S(\alpha)$ sufficiently small.
○ Decompose this cancellation using Vaughan’s identity → Estimate the cancellation using the large sieve inequality, the Cauchy–Schwarz inequality, and related techniques → Derive a result showing that the exponential sum on the minor arcs is much smaller than its trivial upper bound.
○ Conclusion: Estimates for exponential sums in analytic number theory provide an upper bound smaller than the positive main term from the major arcs.
○ Supplement: In the case of the strong Goldbach conjecture, one must consider $\left|S(\alpha)\right|^2$ rather than $\left|S(\alpha)\right|^3$, making it difficult to obtain sufficiently strong upper bounds on the oscillatory contributions. Consequently, the conjecture remains unproven.
⑦ 1-5. Analytically show that $N=\mathfrak{M}\cup\mathfrak{m}$, $R(N)=I_{\mathrm{major}}(N)+I_{\mathrm{minor}}(N)$, and $I_{\mathrm{major}}(N)>\left|I_{\mathrm{minor}}(N)\right|$, thereby establishing $R(N)>0$. $\blacksquare$
⑧ Step 2. Use computer-assisted verification to prove the statement for $7\leq N\leq 8.875694\cdots\times 10^{30}$.
○ Using the result that every even integer up to a certain bound can be expressed as the sum of two primes.
○ Adapting and applying Platt’s verification results.
4. Riemann Hypothesis
⑴ Riemann Hypothesis: Currently unproven.
① Overview
○ Definition: The real part of every nontrivial zero of the Riemann zeta function $\operatorname{Re}(s)=1$ is exactly $1/2$.
○ In the Prime Number Theorem, zeros on the line $\operatorname{Re}(s)=1$ were ruled out. We now aim to extend this zero-exclusion result to a fixed boundary strictly less than 1.
② A error condition equivalent to the Riemann Hypothesis
③ A growth rate condition equivalent to the Riemann Hypothesis
○ To establish a zero-free region for the Riemann zeta function by analytically continuing its reciprocal into the candidate zero region and proving that the continuation is pole-free.
○ Step 1. Choose a conductor $q$: $q=4,5,7,8,9,\cdots$
○ Step 2. Select all primitive characters $\chi$ with that conductor: Construct $L(s,\chi)$ for each character.
○ Step 3. Verify the nontrivial zeros $\rho$ of $L(s,\chi)$: Up to a prescribed height $\left|\operatorname{Im}(\rho)\right|\leq H_q$.
○ Step 4. $\operatorname{Re}(\rho)=1/2$: All nontrivial zeros within the specified range lie on the critical line, namely $\operatorname{Re}(s)=1/2$.
⑵ Quasi-Riemann Hypothesis
① Overview
○ Definition: The Riemann zeta function has no zeros to the right of some fixed boundary strictly less than 1.
○ OpenAI presents two results, Re(𝑠)>11/12 and Re(𝑠)>7/8. However, understanding the former is sufficient to understand a proof of the Quasi-Riemann Hypothesis.
○ Like Goldbach’s conjecture, this is a problem in analytic number theory. However, it had remained unsolved due to the substantially greater complexity and scale of the calculations involved.
② Step 1. Reformulation in terms of the reciprocal of an $L$-function
○ $L(s,\chi)=0 \iff 1/L(s,\chi)$ has a pole.
③ Step 2. Expressing a special number-theoretic sum $J_\eta(Z)$ in two ways
○ Method 1. Direct Estimation: Estimate how small the entire sum is using the theta-function transformation formula, the large sieve inequality, and related techniques.
○ Method 2. Poisson summation formula: $J_\eta(Z)=f_\eta(Z)+E_\eta(Z)$.
○ $f_\eta(Z)$: A Mellin integral associated with $1/L(s,\eta)$.
○ $E_\eta(Z)$: The error term.
④ Step 3. Applying the triangle inequality
○ $\left|f_\eta(Z)\right|\leq \left|J_\eta(Z)\right|+\left|J_\eta(Z)-f_\eta(Z)\right|$.
○ Upper Bound 1: The magnitude of $J_\eta(Z)$ itself is sufficiently small.
○ Upper Bound 2: The difference between $J_\eta(Z)$ and $f_\eta(Z)$ is also sufficiently small.
⑤ Step 4. Analytic continuation via the Mellin transform
○ Applying the Mellin transform allows $1/L(s,\eta)$, which is connected to $f_\eta(Z)$, to be analytically continued to a larger region than originally available.
○ Since $1/L(s,\eta)$ has no poles in this region, it follows that $(L(s,\eta)\neq 0$.
⑥ Step 5. Improving the zero-free boundary
○ Part I: Theta-function transformation formula, Poisson summation formula, and large sieve inequality → No zeros in $\operatorname{Re}(s)>11/12$.
○ Part II: Prime-factor corrections, asymmetric scaling, and additional moment estimates → No zeros in $\operatorname{Re}(s)>7/8$.
Entered: 2026.10.08 00:52