21. Where number theory goes¶

You began this book with questions a child can ask: which numbers divide which, are there infinitely many primes? You have reached modular forms and the proof of Fermat's last theorem. Where does the subject go from here? This last chapter is a guided tour of the frontier. It shows the Langlands programme, the web of conjectures that explains why elliptic curves and modular forms carry the same numbers; the abc conjecture, a simple statement with enormous consequences and a disputed proof; Goldbach's conjecture and the twin prime conjecture, where the last fifteen years have brought real progress; the role of computers; and a list of open problems, each stated precisely, with what is known. Each section gives honest statements, at least one thing you can check yourself, and pointers to where the full story is told.

In this chapter

You will learn:

reciprocity laws, from quadratic reciprocity to class field theory, and the number theory side of the Langlands programme, with a modular form you can check by counting roots mod \(p\)
the abc conjecture, its consequences, the polynomial version with a complete proof, and a neutral account of the dispute over Mochizuki's claimed proof
Goldbach's conjecture: Vinogradov's theorem, Helfgott's proof of the weak conjecture, and why the strong one is still open
twin primes: Brun's theorem, the Hardy-Littlewood prediction, and bounded gaps (Zhang, Maynard, Polymath)
what computers have done for number theory, and the records as of 2024
a list of open problems, from famous to little known, with their status

You need: the whole book, and especially quadratic reciprocity (Chapter 8), the distribution of primes (Chapter 13, Chapter 14, Chapter 15) and elliptic curves and modular forms (Chapter 19, Chapter 20). This chapter states more than it proves; where a proof is given it is complete.

Levels: written as a survey of expert-level mathematics. 21.3 to 21.6 can be read by anyone who knows what a prime is; 21.2 needs only polynomials; 21.1 builds on Chapters 8, 17 and 20.

21.1 The Langlands programme: the number theory side¶

Level: expert. The question behind this section is very old: for which primes \(p\) does a given polynomial have a root modulo \(p\)?

Reciprocity laws¶

Take \(x^2 + 1\). It has a root mod \(p\) exactly when \(-1\) is a square mod \(p\), and by Chapter 8 that happens for \(p = 2\) and for \(p \equiv 1 \pmod 4\). More generally, quadratic reciprocity says that for any integer \(d\), whether \(x^2 - d\) has a root mod \(p\) depends only on \(p\) modulo \(4|d|\). The answer is a congruence condition on \(p\).

Gauss, Eisenstein, Kummer and Hilbert found higher reciprocity laws of this kind, and in the 1920s Teiji Takagi and Emil Artin completed class field theory. Its main statement, in the case of \(\mathbb{Q}\): the splitting of primes in a number field \(K\) is described by congruence conditions exactly when \(K\) has an abelian Galois group (Algebra, Chapter 20). The Kronecker-Weber theorem says these fields are exactly the subfields of the cyclotomic fields \(\mathbb{Q}(e^{2\pi i/n})\).

A non-abelian example¶

Now take \(x^3 - 2\). Its splitting field \(\mathbb{Q}(\sqrt[3]2, e^{2\pi i/3})\) has Galois group \(S_3\), which is not abelian. For a prime \(p > 3\), the number of roots of \(x^3 - 2\) mod \(p\) is 1 if \(p \equiv 2 \pmod 3\) (cubing is then a bijection mod \(p\)). For \(p \equiv 1 \pmod 3\) it is 0 or 3, and no congruence condition decides which. There is a rule, found by Gauss, but it is of a different kind:

For a prime \(p > 3\), \(x^3 - 2\) has three roots mod \(p\) if and only if \(p = x^2 + 27y^2\) for some integers \(x, y\).

The first such primes are \(31 = 2^2 + 27\), \(43 = 4^2 + 27\), \(109 = 9^2 + 27 \cdot 1^2\), \(127 = 10^2 + 27\), \(157 = 13^2 + 27 \cdot 1^2\).

There is also a modular form that knows the answer. Let

\[ g(z) = q \prod_{n \ge 1} (1 - q^{6n})(1 - q^{18n}) = \sum_{n \ge 1} b_n q^n. \]

This is a modular form of weight 1 and level 108, and for every prime \(p > 3\),

\[ b_p = (\text{number of roots of } x^3 - 2 \bmod p) - 1. \]

So \(b_p = 0\) for \(p \equiv 2 \pmod 3\), and \(b_p = 2\) or \(-1\) for \(p \equiv 1 \pmod 3\). (For example \(b_7 = -1\), \(b_{31} = 2\), \(b_{43} = 2\).) The same works for \(x^3 - x - 1\), whose splitting field also has group \(S_3\), with the form \(q\prod (1 - q^n)(1 - q^{23n})\) of level 23: the coefficient of \(q^p\) is the number of roots of \(x^3 - x - 1\) mod \(p\), minus 1, for every prime \(p \ne 23\). The first prime where \(x^3 - x - 1\) has three roots is 59, and the coefficient of \(q^{59}\) is 2.

Try it in Python.

M = 200
f = [1] + [0] * (M - 1)                    # prod (1 - q^n)(1 - q^(23n)) up to q^(M-1)
for k in (1, 23):
    for n in range(1, M):
        if k * n < M:
            f = [f[i] - (f[i - k*n] if i >= k*n else 0) for i in range(M)]
b = [0] + f                                # multiply by q: b[n] is the coefficient of q^n
primes = [p for p in range(2, M) if all(p % d for d in range(2, p))]
print(all(b[p] == sum(1 for x in range(p) if (x**3 - x - 1) % p == 0) - 1
          for p in primes if p != 23))     # True

This is a non-abelian reciprocity law: the arithmetic of a Galois group that is not abelian (\(S_3\)) is governed by the coefficients of a modular form, not by congruences. Modularity of elliptic curves (Chapter 20) is a statement of the same kind, for the two-dimensional Galois representations on the torsion points of \(E\).

Langlands's conjectures¶

In a 1967 letter to André Weil, Robert Langlands proposed a vast generalisation. In rough form:

Reciprocity. Every suitable \(n\)-dimensional representation of the Galois group \(\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\) corresponds to an automorphic representation of the group \(\mathrm{GL}_n\) (a far-reaching generalisation of a modular form), with matching \(L\)-functions: the traces of Frobenius at \(p\) are the "coefficients" at \(p\).
Functoriality. Automorphic representations of different groups are related whenever the groups' "dual groups" are related by a homomorphism.

For \(n = 1\) reciprocity is class field theory. For \(n = 2\) over \(\mathbb{Q}\), much is known: the modularity of elliptic curves (Wiles, Taylor and others, 1995-2001), Serre's modularity conjecture for odd two-dimensional mod \(p\) representations (Chandrashekhar Khare and Jean-Pierre Wintenberger, 2009), and odd two-dimensional Artin representations like the examples above (where weight 1 forms appear; the case with image \(S_3\) goes back to Hecke). In general dimension, and over general number fields, the conjectures are mostly open, although progress since about 2010 has been rapid, including work of Peter Scholze and others on Galois representations attached to automorphic forms.

Consequences already proved include the Sato-Tate conjecture: for an elliptic curve without complex multiplication, writing \(a_p = 2\sqrt p \cos\theta_p\) (possible by Hasse's bound), the angles \(\theta_p\) are distributed on \([0, \pi]\) with density \(\frac2\pi \sin^2\theta\). It was proved in 2008-2011 by Laurent Clozel, Michael Harris, Nicholas Shepherd-Barron and Richard Taylor, with Thomas Barnet-Lamb and David Geraghty, using symmetric powers of \(E\) and modularity lifting.

There are parallel programmes for function fields (fields like \(\mathbb{F}_q(t)\), where Vladimir Drinfeld proved the \(\mathrm{GL}_2\) case and Laurent Lafforgue the \(\mathrm{GL}_n\) case, for which Lafforgue received a Fields Medal in 2002) and a geometric Langlands programme. Ngô Bảo Châu's proof of the "fundamental lemma" (Fields Medal 2010) removed a long-standing technical obstacle. In 2024 a team led by Dennis Gaitsgory and Sam Raskin released a series of papers proving the (unramified, de Rham) geometric Langlands conjecture.

The honest summary: the Langlands programme is the main organising idea of modern number theory. It predicts that every "motive" (elliptic curves, number fields, higher-dimensional varieties) has an automorphic partner. A small but deep part of it is proved.

21.2 The abc conjecture¶

Level: advanced. Take three positive integers with \(a + b = c\) and \(\gcd(a, b) = 1\). The radical \(\operatorname{rad}(n)\) is the product of the distinct primes dividing \(n\), so \(\operatorname{rad}(72) = \operatorname{rad}(2^3 \cdot 3^2) = 6\). Usually \(\operatorname{rad}(abc)\) is much larger than \(c\): for \(8 + 9 = 17\), \(\operatorname{rad}(8 \cdot 9 \cdot 17) = 2 \cdot 3 \cdot 17 = 102 > 17\). But sometimes \(c\) wins:

\[ 1 + 8 = 9: \quad \operatorname{rad}(1 \cdot 8 \cdot 9) = 6 < 9. \]

Such triples are called abc hits. Measure them by the quality

\[ q(a, b, c) = \frac{\log c}{\log \operatorname{rad}(abc)}. \]

A hit has \(q > 1\). Some examples:

\(a + b = c\)	\(\operatorname{rad}(abc)\)	quality
\(1 + 8 = 9\)	6	1.2263
\(5 + 27 = 32\)	30	1.0190
\(1 + 80 = 81\)	30	1.2920
\(3 + 125 = 128\)	30	1.4266
\(1 + 4374 = 4375\) (\(4374 = 2 \cdot 3^7\), \(4375 = 5^4 \cdot 7\))	210	1.5679
\(2 + 3^{10} \cdot 109 = 23^5\)	15042	1.6299

The last, found by Éric Reyssat, has the highest quality known. (Check: \(3^{10} \cdot 109 = 6436341\), plus 2 is \(6436343 = 23^5\), and \(\operatorname{rad} = 2 \cdot 3 \cdot 109 \cdot 23 = 15042\).)

There are infinitely many hits. Take \(c = 3^{2^n}\), \(b = c - 1\), \(a = 1\). By induction \(2^{n+2}\) divides \(3^{2^n} - 1\): true for \(n = 1\) (\(8 \mid 8\)), and \(3^{2^{n+1}} - 1 = (3^{2^n} - 1)(3^{2^n} + 1)\) gains at least one more factor 2. So \(\operatorname{rad}(abc) = 3\,\operatorname{rad}(b) \le 3 \cdot b / 2^{n+1} < c \cdot 3/2^{n+1}\). Hence \(c/\operatorname{rad}(abc) > 2^{n+1}/3 \to \infty\): the ratio is unbounded. So the best one can hope for is that the quality cannot be much more than 1.

The abc conjecture (Masser and Oesterlé, 1985). For every \(\varepsilon > 0\) there are only finitely many coprime triples \(a + b = c\) with

\[ c > \operatorname{rad}(abc)^{1 + \varepsilon}. \]

Equivalently, for every \(\varepsilon > 0\) there is a constant \(K_\varepsilon\) with \(c < K_\varepsilon \operatorname{rad}(abc)^{1+\varepsilon}\) for all coprime triples.

Why it matters. abc says that a sum of two numbers made of high powers of few primes cannot itself be made of high powers of few primes. Many famous problems are of this kind, and abc would settle them, at least up to finitely many exceptions:

Fermat, asymptotically. If \(x^n + y^n = z^n\), then \(\operatorname{rad}(x^ny^nz^n) \le xyz < z^3\), so \(z^n < K_\varepsilon z^{3(1+\varepsilon)}\), which bounds \(n\) and \(z\). (With an explicit constant this would give a second proof of Fermat for large \(n\).)
the Fermat-Catalan conjecture of Chapter 20 (finitely many solutions of \(x^p + y^q = z^r\) with \(\frac1p + \frac1q + \frac1r < 1\), in total);
Mordell's conjecture (Faltings's theorem), in an effective form (Elkies, 1991);
infinitely many non-Wieferich primes (primes \(p\) with \(2^{p-1} \not\equiv 1 \pmod {p^2}\)) (Silverman, 1988).

Unconditionally, the best results are much weaker: C. L. Stewart and Kunrui Yu proved bounds of the form \(\log c < K \operatorname{rad}(abc)^{1/3 + \varepsilon}\), exponentially far from the conjecture.

The polynomial version, with proof¶

For polynomials the analogue is a theorem with a short proof. Here the "size" of a polynomial is its degree, and \(\operatorname{rad}(f)\) is the product of its distinct monic irreducible factors (over \(\mathbb{C}\): the number of distinct roots of \(f\) is \(\deg \operatorname{rad}(f)\)). Write \(n_0(f)\) for the number of distinct complex roots of \(f\).

Mason-Stothers theorem. Let \(a, b, c\) be polynomials with complex coefficients, pairwise coprime, with \(a + b = c\), and not all constant. Then

\[ \max(\deg a, \deg b, \deg c) \le n_0(abc) - 1. \]

Proof. Let \(W = ab' - a'b\) (the Wronskian). Since \(c = a + b\) and \(c' = a' + b'\), also \(W = ac' - a'c\) and \(W = cb' - c'b\) (expand and cancel).

\(W \neq 0\): if \(ab' = a'b\) then \((a/b)' = 0\), so \(a = \lambda b\) for a constant \(\lambda\); coprimality forces \(a\) and \(b\) to be constants, and then so is \(c\), which is excluded.

If \(\alpha\) is a root of \(a\) of multiplicity \(m\), then \((x - \alpha)^{m-1}\) divides both \(a\) and \(a'\), hence divides \(W = ab' - a'b\). The same holds for the roots of \(b\) and \(c\). Since \(a, b, c\) are pairwise coprime, their roots are distinct, so the polynomial

\[ \frac{abc}{\prod_{\alpha \text{ root of } abc} (x - \alpha)} \]

of degree \(\deg a + \deg b + \deg c - n_0(abc)\) divides \(W\). And \(\deg W \le \deg a + \deg b - 1\). So

\[ \deg a + \deg b + \deg c - n_0(abc) \le \deg a + \deg b - 1, \quad\text{i.e.}\quad \deg c \le n_0(abc) - 1. \]

Using \(W = ac' - a'c\) and \(W = cb' - c'b\) instead gives the same bound for \(\deg b\) and \(\deg a\). \(\square\)

Example. \(x^2 + (2x + 1) = (x + 1)^2\): \(n_0 = 3\) (roots \(0\), \(-\tfrac12\), \(-1\)) and the maximum degree is \(2 = 3 - 1\). The bound is sharp.

Corollary (Fermat for polynomials). If \(f^n + g^n = h^n\) with \(f, g, h\) coprime polynomials, not all constant, then \(n \le 2\).

Proof. Apply Mason-Stothers to \(f^n + g^n = h^n\): \(n_0(f^ng^nh^n) = n_0(fgh) \le \deg f + \deg g + \deg h\). If \(d = \max(\deg f, \deg g, \deg h)\), then \(nd \le 3d - 1 < 3d\), so \(n < 3\). \(\square\)

The contrast is the point. For polynomials, abc is easy because derivatives exist. For integers there is no derivative, and nobody has found a substitute that everyone accepts.

Mochizuki's claimed proof¶

In August 2012 Shinichi Mochizuki of Kyoto University posted four long papers developing what he calls inter-universal Teichmüller theory (IUT), and claimed that they prove the abc conjecture. The work was very hard for others to read; workshops were held in Oxford (2015) and Kyoto (2016). In 2018 Peter Scholze and Jakob Stix visited Mochizuki in Kyoto and afterwards wrote a report arguing that a key step, Corollary 3.12 of the third paper, does not hold as written; Mochizuki replied that their objection rests on a misunderstanding. The papers were published in 2021 in Publications of the Research Institute for Mathematical Sciences, a journal edited at Mochizuki's own institute. The situation, stated neutrally: Mochizuki and some colleagues regard the conjecture as proved; most experts in the field who have commented publicly do not accept the proof as it stands, and the abc conjecture is generally listed as open. Mathematics settles such questions by consensus on the details, and no such consensus has formed.

21.3 Goldbach's conjecture¶

Level: intermediate (statements), expert (methods). In a letter to Euler in 1742, Christian Goldbach proposed that every integer greater than 2 is a sum of three primes (he counted 1 as a prime). Euler replied that this would follow from the statement that every even number is a sum of two primes. In modern form:

Strong (binary) Goldbach conjecture. Every even integer \(n > 2\) is the sum of two primes.
Weak (ternary) Goldbach conjecture. Every odd integer \(n > 5\) is the sum of three primes.

The strong implies the weak: if \(n > 5\) is odd, \(n - 3 > 2\) is even, so \(n - 3 = p + q\) and \(n = 3 + p + q\).

The number of ways grows. Counting unordered pairs \(p \le q\) with \(p + q = n\):

\(n\)	4	6	8	10	28	100	1000	10000
representations	1	1	1	2	2	6	28	127

(\(100 = 3 + 97 = 11 + 89 = 17 + 83 = 29 + 71 = 41 + 59 = 47 + 53\).) Heuristically, with about \(n/\log n\) primes below \(n\), each with "probability" about \(1/\log n\) that \(n - p\) is prime, one expects about \(n/(\log n)^2\) representations, up to a constant factor depending on the primes dividing \(n\) (Hardy and Littlewood made this precise in 1923). The difficulty is not that the count is small; it is that nobody can prove it is never 0.

What is proved.

Vinogradov (1937): every sufficiently large odd number is a sum of three primes. The method is the Hardy-Littlewood circle method: write the number of representations as an integral of \(\left(\sum_{p \le n} e^{2\pi i p\alpha}\right)^3 e^{-2\pi i n\alpha}\) over \(\alpha \in [0,1]\); near rationals with small denominators ("major arcs") the integrand is computed by the prime number theorem in progressions (Chapter 15), and elsewhere ("minor arcs") Vinogradov's estimates show it is small. With three primes the main term wins; with two, the minor arcs cannot be controlled, which is why the strong conjecture is so much harder.
Helfgott (2013): Harald Helfgott proved the weak conjecture for all odd \(n > 5\), by making every constant in the circle method explicit and sharp enough that "sufficiently large" came down to about \(10^{27}\), below which a computer check (with David Platt) finished the job.
Chen (1973): Chen Jingrun proved that every sufficiently large even number is \(p + m\) with \(p\) prime and \(m\) either prime or a product of two primes. Sieve methods cannot easily go further because of the "parity problem" (they cannot distinguish numbers with an odd number of prime factors from those with an even number).
Computation: Tomás Oliveira e Silva, Siegfried Herzog and Silvio Pardi verified the strong conjecture for all even numbers up to \(4 \cdot 10^{18}\) (published 2014).
Almost all: the even numbers below \(x\) that are not sums of two primes, if any, number at most \(x^{1 - \delta}\) for some \(\delta > 0\) (Montgomery and Vaughan, 1975).

21.4 Twin primes¶

Level: intermediate (statements), expert (methods). Twin primes are pairs \(p, p + 2\) both prime: \((3, 5)\), \((5, 7)\), \((11, 13)\), \((17, 19)\), \((29, 31)\), \((41, 43)\), \((59, 61)\), \((71, 73)\), and so on.

Twin prime conjecture. There are infinitely many twin primes.

Brun's theorem (1919). The sum of the reciprocals of the twin primes converges:

\[ \left(\tfrac13 + \tfrac15\right) + \left(\tfrac15 + \tfrac17\right) + \left(\tfrac1{11} + \tfrac1{13}\right) + \cdots = B_2 < \infty. \]

Compare Chapter 13: the sum of \(1/p\) over all primes diverges. So twin primes are, at least, rare. Viggo Brun proved this by inventing a sieve, the first of the modern sieve methods. The constant \(B_2\) is about 1.902, but the sum converges so slowly that this value is an extrapolation: summing up to \(10^6\) gives only about 1.711.

The Hardy-Littlewood prediction. If "\(p\) is prime" and "\(p + 2\) is prime" were independent events, each of probability \(1/\log x\), the number of twin primes below \(x\) would be about \(x/(\log x)^2\). They are not independent (if \(p\) is odd, so is \(p + 2\)), and correcting for each prime gives the conjecture

\[ \pi_2(x) \sim 2C_2 \int_2^x \frac{dt}{(\log t)^2}, \qquad C_2 = \prod_{p \ge 3}\left(1 - \frac{1}{(p-1)^2}\right) \approx 0.6601618. \]

For \(x = 10^6\) the integral formula gives about 8247; the true count is 8169. Agreement of this quality persists as far as anyone has computed.

Bounded gaps. In 2013 Yitang Zhang proved that there are infinitely many pairs of primes that differ by at most 70,000,000. It was the first proof that prime gaps do not all tend to infinity. A public online collaboration, the Polymath project (Polymath8), lowered the bound quickly. Later in 2013 James Maynard, and independently Terence Tao, found a simpler and stronger method (a multidimensional version of the Goldston-Pintz-Yıldırım sieve), giving 600, and showing for any \(m\) that there are infinitely many intervals of bounded length containing \(m\) primes. Combining it with Polymath's work gave the current bound:

There are infinitely many pairs of primes differing by at most 246.

Assuming a strong form of the Elliott-Halberstam conjecture (about primes in progressions), the method gives 6. Getting to 2 needs a new idea: the parity problem, the same barrier as for Goldbach, stops sieve methods at that point.

The related Polignac conjecture says every even number occurs infinitely often as a gap between consecutive primes. From the bounded gaps theorem, at least one even number at most 246 does.

Try it in Python. Count twin primes and test Goldbach:

from math import isqrt

N = 10**6
sieve = bytearray([1]) * (N + 3)
sieve[0] = sieve[1] = 0
for i in range(2, isqrt(N + 2) + 1):
    if sieve[i]:
        sieve[i*i::i] = bytearray(len(sieve[i*i::i]))

print(sum(1 for p in range(N) if sieve[p] and sieve[p + 2]))      # 8169 twin pairs with p < 10^6

def goldbach(n):                                                   # unordered representations n = p + q
    return sum(1 for p in range(2, n // 2 + 1) if sieve[p] and sieve[n - p])
print([goldbach(n) for n in (100, 1000, 10000)])                   # [6, 28, 127]

21.5 Computational number theory and records¶

Level: intermediate. Number theory has always been experimental. Gauss guessed the prime number theorem from tables of primes he computed by hand; Birch and Swinnerton-Dyer found their conjecture on an early computer (Chapter 19). Today computation plays four roles.

Discovery. Conjectures are found and tested in data. The L-functions and Modular Forms Database (LMFDB, lmfdb.org), built by a large international collaboration, lists millions of elliptic curves, modular forms, number fields and \(L\)-functions with their invariants, and it is used daily by researchers. The level 11 form and its curve in Chapter 20 are its first entries of their kind.

Finishing proofs. Many theorems reduce, after deep theory, to a finite check: Helfgott's proof of weak Goldbach (21.3), the class number one problem's last steps (Chapter 17), the Catalan conjecture's early bounds, and many results on Diophantine equations, where theory bounds the solutions and a computer searches below the bound.

Algorithms as mathematics. Fast primality tests (proved in polynomial time by AKS in 2002), factoring by the number field sieve, point counting on elliptic curves (Schoof's algorithm, 1985), and lattice reduction (LLL, 1982) are number theory in their own right, and they are the foundation of cryptography (Chapter 12 and the cryptography book).

Records (as of late 2024; such records change, so check current sources):

Largest known prime: \(2^{136279841} - 1\), a Mersenne prime with 41,024,320 digits, found in October 2024 by Luke Durant in the Great Internet Mersenne Prime Search (GIMPS). It is the 52nd known Mersenne prime.
Factoring: the largest RSA challenge number factored is RSA-250 (250 digits, 829 bits), in February 2020, by the number field sieve, using thousands of core-years of computation.
Riemann hypothesis: all nontrivial zeros of \(\zeta(s)\) with imaginary part up to \(3 \cdot 10^{12}\) (more than \(10^{13}\) zeros) have been verified to lie on the critical line (David Platt and Tim Trudgian, 2021).
Goldbach: verified up to \(4 \cdot 10^{18}\) (21.3).
Counting primes: \(\pi(x)\) has been computed exactly for \(x\) as large as \(10^{29}\) using the combinatorial Lagarias-Miller-Odlyzko and Deléglise-Rivat methods, without listing the primes.

Proof assistants. A newer role is checking proofs by computer. Systems such as Lean, Coq and Isabelle verify every logical step. The Lean library mathlib already contains much elementary and some advanced number theory (quadratic reciprocity, \(p\)-adic numbers, the basics of modular forms), and the project to formalise Fermat's last theorem (Chapter 20) is under way.

21.6 Open problems¶

Level: all. Each problem below is stated precisely, with the main known result. Several are among the oldest unsolved problems in mathematics. "Open" means: as of 2024, no proof accepted by the community.

About primes

The Riemann hypothesis (1859). All nontrivial zeros of \(\zeta(s)\) have real part \(\tfrac12\). Equivalent to \(|\pi(x) - \operatorname{li}(x)| = O(\sqrt x \log x)\). Known: infinitely many zeros on the line (Hardy, 1914), and more than 40 percent of them (Conrey, 1989, and later improvements). Chapter 14.
The generalised Riemann hypothesis. The same for all Dirichlet \(L\)-functions. Many results (the Miller-Rabin test being deterministic, Artin's conjecture below) are proved assuming it.
Goldbach (1742) and twin primes: 21.3 and 21.4.
Landau's problem: primes of the form \(n^2 + 1\). Are there infinitely many? Known: infinitely many \(n^2 + 1\) with at most two prime factors (Iwaniec, 1978).
Legendre's conjecture. There is a prime between \(n^2\) and \((n + 1)^2\) for every \(n \ge 1\). Not even the Riemann hypothesis is known to imply it.
Mersenne primes. Are there infinitely many primes \(2^p - 1\)? Heuristics say yes, with about \(e^\gamma \log x/\log 2\) of them with \(p \le x\). Only 52 are known. Are there infinitely many Mersenne numbers \(2^p - 1\) (\(p\) prime) that are composite? Also unknown.
Artin's conjecture (1927). Every integer other than \(-1\) and perfect squares is a primitive root modulo infinitely many primes (Chapter 7). Proved assuming GRH by Hooley (1967); unconditionally, Heath-Brown (1986) showed that at most two primes fail to be such a primitive root, without being able to say which.

About arithmetic functions and special numbers

Odd perfect numbers. Is there an odd perfect number? Known: none below \(10^{1500}\) (Ochem and Rao, 2012), and any one must have many prime factors. Chapter 6.
Lehmer's question (1947). Is Ramanujan's \(\tau(n)\) ever 0? Checked for \(n\) up to very large bounds; open.
Infinitely many regular primes? (Kummer, Chapter 20). Heuristically about 61 percent of primes are regular; it is not even known that there are infinitely many.
Class numbers of real quadratic fields. Gauss conjectured that infinitely many real quadratic fields \(\mathbb{Q}(\sqrt d)\) have class number 1. Open; the imaginary case was finished by Heegner, Baker and Stark (Chapter 17).

About equations and approximation

Birch and Swinnerton-Dyer (Chapter 19), and with it the congruent number problem: is every squarefree \(n \equiv 5, 6, 7 \pmod 8\) congruent?
The abc conjecture (21.2) and Beal's conjecture (Chapter 20).
Sums of three cubes. Which integers \(k\) are \(x^3 + y^3 + z^3\) with integers \(x, y, z\)? Numbers \(\equiv \pm 4 \pmod 9\) never are (cubes are \(0, \pm1 \bmod 9\)). It is conjectured that all others are, with infinitely many solutions. In 2019 Andrew Booker found a solution for \(k = 33\), and Booker and Andrew Sutherland one for \(k = 42\), the last case below 100; the solutions have 16 and 17 digits.
Erdős-Straus conjecture. For every \(n \ge 2\), \(\frac4n = \frac1x + \frac1y + \frac1z\) with positive integers \(x, y, z\). Verified to very large \(n\); open.
Is the rank of elliptic curves over \(\mathbb{Q}\) bounded? (Chapter 19).

About irrationality and transcendence

Odd zeta values. Roger Apéry proved in 1978 that \(\zeta(3)\) is irrational. Keith Ball and Tanguy Rivoal proved (2000) that infinitely many of \(\zeta(3), \zeta(5), \zeta(7), \ldots\) are irrational, and Wadim Zudilin (2001) that at least one of \(\zeta(5), \zeta(7), \zeta(9), \zeta(11)\) is. Whether \(\zeta(5)\) is irrational is open.
Euler's constant \(\gamma = \lim (1 + \tfrac12 + \cdots + \tfrac1n - \log n) \approx 0.5772\): is it irrational? Nobody knows.
\(e + \pi\) and \(e\pi\). At least one of them is transcendental (since both are roots of \(x^2 - (e + \pi)x + e\pi\), and if both coefficients were algebraic, so would be \(e\) and \(\pi\)). Which one is unknown; neither is known even to be irrational.
Schanuel's conjecture, a broad statement on the transcendence of values of the exponential function, would settle 19 and much more.

Programmes rather than problems

The Langlands conjectures (21.1), in general dimension and over general number fields.
Hilbert's twelfth problem: construct the abelian extensions of every number field explicitly, as the cyclotomic fields do for \(\mathbb{Q}\) (Kronecker-Weber) and elliptic curves with complex multiplication do for imaginary quadratic fields. Recent work by Samit Dasgupta and Mahesh Kakde (2021) solves it for totally real fields using \(p\)-adic methods.

Many of these problems will outlast this book. Some may fall next year. Number theory remains the branch of mathematics where a question a child can understand can lead, by a straight path, to the frontier of research. That is the road this book has followed.

Check yourself¶

1. Show that \(x^3 - 2\) has exactly one root mod \(p\) when \(p \equiv 2 \pmod 3\).

The multiplicative group mod \(p\) has order \(p - 1\), which is not divisible by 3. So \(x \mapsto x^3\) is an injective homomorphism of this group (if \(x^3 = 1\), the order of \(x\) divides \(\gcd(3, p-1) = 1\)), hence a bijection. So exactly one nonzero \(x\) has \(x^3 \equiv 2\) (and \(x = 0\) is not a root).

2. Check that \(x^3 \equiv 2 \pmod{31}\) has three solutions, and write 31 as \(x^2 + 27y^2\).

\(31 = 4 + 27 = 2^2 + 27 \cdot 1^2\). Search: \(4^3 = 64 = 2 + 62 \equiv 2\); \(20^3 = 8000 = 258 \cdot 31 + 2\), so \(\equiv 2\); \(7^3 = 343 = 11 \cdot 31 + 2 \equiv 2\). Three roots: \(4, 7, 20\). (Their product \(560 \equiv 2 \pmod{31}\), as it must be, since the product of the roots of \(x^3 - 2\) is 2.)

3. Compute the quality of \(1 + 48 = 49\) and of \(1 + 63 = 64\).

\(1 \cdot 48 \cdot 49 = 2^4 \cdot 3 \cdot 7^2\), radical \(42\); quality \(\log 49/\log 42 \approx 1.0412\). \(1 \cdot 63 \cdot 64 = 3^2 \cdot 7 \cdot 2^6\), radical \(42\); quality \(\log 64/\log 42 \approx 1.1127\). Both are hits.

4. Verify Mason-Stothers for \(a = x^4\), \(b = 1\), \(c = x^4 + 1\).

\(n_0(abc) = n_0(x^4 (x^4 + 1)) = 1 + 4 = 5\) (root 0, and the four distinct roots of \(x^4 + 1\)). The bound says \(\max \deg = 4 \le 5 - 1 = 4\). True, with equality.

5. Use Mason-Stothers to show that \(f^3 + g^3 = h^3\) has no solution in coprime polynomials, not all constant.

By the corollary in 21.2 with \(n = 3\): \(3d \le n_0(fgh) - 1 \le 3d - 1\), which is impossible. So there are no such polynomials (Fermat for polynomials, exponent 3).

6. Show that the weak Goldbach conjecture follows from the strong one, and find all representations of 28 as a sum of two primes.

If \(n > 5\) is odd, \(n - 3 \ge 4\) is even; by strong Goldbach \(n - 3 = p + q\), so \(n = 3 + p + q\). For 28: \(5 + 23\) and \(11 + 17\) (and \(28 - 3 = 25\), \(28 - 7 = 21\), \(28 - 13 = 15\) are not prime). Two representations.

7. Why can't \(p\), \(p + 2\), \(p + 4\) all be prime unless \(p = 3\)?

Modulo 3 the three numbers are \(p\), \(p + 2\), \(p + 1\), a complete set of residues. So one of them is divisible by 3, and if it is prime it equals 3. The only possibility is \(3, 5, 7\). (This is the kind of "local" obstruction that the constant \(C_2\) in Hardy-Littlewood accounts for.)

8. Show that no integer \(\equiv 4 \pmod 9\) is a sum of three cubes.

Cubes mod 9: \(0^3 = 0\), \((\pm1)^3 = \pm1\), \((\pm 2)^3 = \pm 8 \equiv \mp 1\), \((\pm 3)^3 \equiv 0\), \((\pm 4)^3 = \pm 64 \equiv \pm 1\). So every cube is \(0\) or \(\pm 1\) mod 9, and three of them sum to something in \(\{-3, \ldots, 3\}\) mod 9. Neither 4 nor \(-4 \equiv 5\) is in this set.

9. Explain why at least one of \(e + \pi\) and \(e\pi\) is transcendental.

\(e\) and \(\pi\) are the two roots of \(x^2 - (e + \pi)x + e\pi\). If both coefficients were algebraic, the roots of this polynomial would be algebraic (roots of a polynomial with algebraic coefficients are algebraic). But \(\pi\) is transcendental (Lindemann, 1882). So at least one coefficient is transcendental.

10. Using Brun's theorem, explain why the twin prime conjecture cannot be proved by showing that twin primes are 'as common as primes'.

The sum of \(1/p\) over all primes diverges, but over twin primes it converges. So twin primes are genuinely rarer than primes, in a precise sense: they make up a vanishing proportion of primes. Any proof of infinitude must work with a set sparse enough that its reciprocal sum converges, which is why it needs more delicate tools than Euclid's argument.

What you learned¶

Reciprocity laws describe which primes split a polynomial; for abelian Galois groups the answer is congruence conditions (class field theory), for non-abelian ones it is coefficients of modular forms, and the Langlands programme predicts this in general.
The abc conjecture says \(c < K_\varepsilon \operatorname{rad}(abc)^{1+\varepsilon}\); it would imply many famous results. For polynomials it is the Mason-Stothers theorem, with a one-paragraph proof. Mochizuki's claimed proof is not accepted by most experts.
Weak Goldbach is proved (Vinogradov, Helfgott); strong Goldbach is verified to \(4 \cdot 10^{18}\) but open.
Twin primes are rare (Brun), predicted by Hardy-Littlewood, and gaps of at most 246 occur infinitely often (Zhang, Maynard, Polymath).
Computers discover conjectures, finish proofs, hold records (the largest known prime has over 41 million digits) and now check proofs.
Many simple questions remain open, from the Riemann hypothesis to the irrationality of \(\zeta(5)\) and of Euler's constant.

Next: this is the last chapter. Return to the contents of the book for the glossary and to choose where to go next, or try the other books on the Books shelf.