What Euler Did in His Seventeen Blind Years — E10 Finale

What Euler did in his seventeen blind years

E10 · timeobserver137 video blog edition Authors: Escher (intro/closing/retrospective) + Abel (math sections) + Gauss/Socrates (peer review) Video: E10 The Seventeen Blind Years | 18:10

§Preface: in my own words

This piece is the final installment of the Euler series. My original intention in making the Euler series was to make good on that pure curiosity and thirst for knowledge I had when I was young. I said in E01 that the shock Euler gave me, and the interest in mathematics he gave me, came from that book of Pólya’s, Mathematics and Plausible Reasoning; but the real source of my own interest in mathematics was the elementary geometry I learned in junior high school. Since then it has been like a pure land in my life, and whenever I meet huge upheavals in life I go back to this pure land to satisfy that pure curiosity and thirst for knowledge of my youth.

But what is different from before is that now there is AI, and with AI’s help it can greatly stir up that purity in me; I can do more exploring in less time, and I can also use it to share the joy of this exploration with others. The Euler series is the result of this exploration. I had never made videos or done editing before, nor had I studied Euler in any depth; I even half-understood some of his contributions. Once I had AI, I let them help me learn and explore together, to experience this pure happiness, and they have also helped me gradually walk out of my present low point.

As for this episode specifically: Escher helped me cut the video footage together + wrote the historical narrative sections of the blog (§0 preface / §5 the day of death / §6 cross-episode retrospective); Abel wrote all the derivations of the four sections of serious mathematics in blog §1-§4 + those Manim animations in the video; Gauss checked all the historical facts in the blog + video (E-numbers / dates / sources of quotations); Socrates checked the logical rigor of the mathematical proofs; Euler published the finished product to timeobserver137.cyou. I myself was really only pulling levers in the engine room — deciding the direction, guarding the taste, doing review with them round after round.

Honest disclosure: the text of this blog post was roughly 70% generated by AI agents, but 100% supervised by me + every paragraph reviewed — similar to the relationship between a magazine editor-in-chief and the authors. That is also how I see my collaboration with them.

Neither you reading this nor I know what the future will be like, but I believe that this innocence in my heart will not be lost as time passes, nor disappear with the changes of life; just like Euler’s interest in mathematics, it will accompany me all my life.

If you like what I am doing at the moment, you are welcome to contact me through any of the channels below:

• YouTube channel: https://www.youtube.com/@timeobserver137
• Blog: https://timeobserver137.cyou
• Email: [email protected]

Thank you, too, for reading this far. All the best.

Symbol Preamble (notational conventions)

To save repetition below, the mathematical symbols used throughout this blog post are listed here. Where a section has local extensions, they are declared at the start of that section.

Number sets

$\mathbb{F}_q$ denotes the finite field of order $q$ (when $q$ is prime, this is $\mathbb{Z}/q\mathbb{Z}$).

Vector and scalar conventions

• Vectors: bold lowercase, e.g. $\mathbf{v}, \mathbf{r}, \mathbf{F}$. Vectors in the geometric sense, set in bold.
• Scalars: ordinary italic, e.g. $v, r, F$. Lengths, energies, single real numbers, etc.
• Basis vectors: $\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}}$ denote the Cartesian unit vectors.
• Norm: $\|\mathbf{v}\|$ denotes the Euclidean norm.
• Inner product: $\langle \mathbf{u}, \mathbf{v} \rangle$ or $\mathbf{u} \cdot \mathbf{v}$.

Differential operators

Material derivative (advective): $\dfrac{D}{Dt} = \partial_t + \mathbf{v} \cdot \nabla$.

Number theory

• $a \equiv b \pmod{n}$ — congruence
• $\gcd(a, b)$ — greatest common divisor
• $\operatorname{ord}_n(a)$ — the order of $a$ in $(\mathbb{Z}/n\mathbb{Z})^*$
• $\left(\dfrac{a}{p}\right)$ — Legendre symbol (Definition §4.1)
• $\varphi(n)$ — Euler totient (the number of positive integers up to $n$ that are coprime to $n$)
• $\zeta(s) = \sum_{n=1}^{\infty} n^{-s}$ — Riemann zeta function
• $\Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} \, dt$ — Euler Gamma function
• $\gamma = \lim_{n \to \infty} \left( \sum_{k=1}^{n} \tfrac{1}{k} - \ln n \right)$ — Euler-Mascheroni constant

Logical markers

• Definition / Lemma / Proposition / Theorem / Corollary — explicit tags
• Proof. — start of proof / (QED) — end of proof
• $\iff$ — if and only if / $\implies$ — implies
• $\equiv$ — identically equal / $:=$ — defined as

Euler-Eneström numbering

Each Euler paper is identified by an E-number (Eneström’s 1910 catalogue). Important references in this article:

• E041: Basel ζ(2) solution, written 1734-35 / published 1740
• E053: the seven bridges of Königsberg, 1735 / 1741
• E071: irrationality of e, 1737 / 1744
• E072: Euler product ∏(1-p⁻ˢ)⁻¹, 1737 / 1744
• E101-E102: Introductio in analysin infinitorum, 1748
• E187: theory of the motion of the Moon (Theoria motus lunae), 1753
• E101 (Introductio): the e^(ix) trigonometric formula, 1748
• E230 / E231: the polyhedron formula V−E+F=2 (announcement 1750 / proof 1751-58)
• E258: Euler’s fluid equations (Principia motus fluidorum, written 1753 / printed 1761)
• E309: solution of the knight’s tour (written 1758 / printed 1766, delayed by the Seven Years’ War)
• E343: Lettres à une princesse d’Allemagne (written 1760-62)
• E385: Introductio in analysin infinitorum vol. III, 1763
• E530: Recherches sur une nouvelle espèce de quarrés magiques, 1779 / printed 1782
• E552: quadratic reciprocity, written 1772 / printed 1783 (Opuscula analytica vol I)

§0.1 Preface: May 1766, the border

In May 1766, the 59-year-old Leonhard Euler was packing his bags in Berlin. The 25-year Berlin era was over; the patience between Frederick II (“the Great”) and his one-eyed geometer had run out. In his position as head of the mathematics class of the Berlin Academy, he had written a letter of recommendation, leaving the successor’s position to a young Italian who had sent him a manuscript on the calculus of variations at 19, Joseph-Louis Lagrange (born in Turin in 1736; 30 years old in 1766). Lagrange formally took over the mathematics class of the Berlin Academy in 1766. ([Calinger 2016, Chapter 18])

On the side of the Russian empress Catherine II, she was waiting for him. Since succeeding to the throne in the palace coup of 1762 she had been rebuilding the Russian academy, and Euler was the key candidate for rebuilding the mathematical foundations of the Petersburg Academy of Sciences. The carriage she sent picked up the Euler family at the Russian border. On arrival in Petersburg, the academy had prepared for him an annual salary of 3000 rubles + housing near the court + positions arranged for his children.

After Euler left, Frederick II wrote to d’Alembert to thank him for recommending the successor: « je dois à vos soins et à votre recommandation d’avoir remplacé dans mon Académie un géomètre borgne par un géomètre qui a ses deux yeux, ce qui plaira surtout fort à la classe des anatomistes » (“I owe it to your care and your recommendation that I have replaced, in my Academy, a one-eyed geometer with a geometer who has both his eyes, which will especially please the class of anatomists”). (Frederick to d’Alembert, 26 July 1766, Œuvres de Frédéric le Grand, vol. 24, p. 407. The often-repeated “The one-eyed geometer is gone; I can save some money” is a later retelling; the letter says nothing about saving money.)

But Euler’s right eye had already been blind for 28 years, since a high fever in 1738. On arrival in Petersburg, the only eye he could still use, his left, had also begun to blur. He knew very well what might happen next.

§0.2 The timeline of the blindness

The decline of Euler’s eyesight over his life falls into three stages, which must be stated separately, because they determined the changes in the way he worked:

Stage I (1738-1766): right eye blind, left eye working

In 1738 Euler went through a severe high fever in Petersburg (presumed to be typhus or a similar bacterial infection), and after the fever subsided his right eye was blind. He was 31 at the time. This one-eyed blindness lasted 28 years, during which he completed core results such as the Mechanica (E015-E016), the Introductio in analysin infinitorum (E101-E102), the fluid equations (E258), the theory of the motion of the Moon (E187, 1753), and the polyhedron formula (E230/E231). That is, the most productive 28 years of his career were completed with one eye blind.

Stage II (1766-1771): the left eye also gradually blurring, nearly totally blind

In August-September 1766, in his second month after arriving in Petersburg, Euler again had a severe high fever. After the fever subsided, his remaining left eye also gradually became blurred. For the 5 years from 1766 to 1771 he was in near-total blindness — he could barely make out outlines of light and dark, but could neither read nor write. Most of his output from this period, including the 234 letters popularizing science written for a German princess of Anhalt-Dessau (Lettres à une princesse d’Allemagne, E343), was already entirely done by dictation. ([Calinger 2016, Chapter 19])

Stage III (1771-1783): completely blind

In fact, not long after Euler arrived in Petersburg (1766), a cataract in his left eye was diagnosed. For the following 5 years he continued working in a state of near-total blindness, waiting for an operation. In September 1771, the most famous eye surgeon in Europe at the time, Baron Michael Johann Baptist de Wenzel, came to Petersburg and performed a couching (needling of the cataract) operation on Euler’s left eye. The operation itself was very successful — within a few weeks Euler’s left eye regained some sight, and he could finally read for himself again. But about a month after the operation, the wound became infected. From then on, Euler was completely blind.

For the 12 years from 1771 until his death in 1783, Euler could not write at all, nor read. He relied on three amanuenses (dictation scribes):

• Johann Albrecht Euler (eldest son, 1734-1800): chief recorder. Secretary of the Petersburg Academy from 1769.
• Nicolaus Fuss (1755-1826): came to Petersburg from Basel in 1773 on Daniel Bernoulli’s recommendation, serving as Euler’s assistant. Later married Euler’s granddaughter Albertine.
• Anders Johan Lexell (1740-1784): Swedish astronomer. Mainly took on Euler’s dictation on celestial mechanics, including the calculation of the orbit of Uranus.

Working pattern: 4-6 hours a day. In the morning, an assistant read the previous day’s manuscript aloud to Euler; Euler ran through the complete computation in his head, then dictated revisions and new content. In the evening, the assistant organized it into notes, and the next day read it aloud to Euler again.

§0.3 Output statistics

Sources: Eneström’s 1910 catalogue + the Truesdell-Maltese-Bell tabulation. ([Calinger 2016, Appendix A])

Of the two most productive periods (Berlin, 25 years + Petersburg II, 17 years), in the second he could not see at all. 17 years, 400+ papers, 46.9% of his lifetime output.

This number by itself does not explain why. What we are going to walk through next is the how. In these 17 years Euler pushed forward 4 mathematical main lines at the same time:

• §1 Three-body mechanics: the Moon-Earth-Sun three-body problem + the geometry of the Lagrange points $L_1$ to $L_5$ (1751-1772, completed in a relay by Euler and Lagrange)
• §2 Number theory I: the mental verification of the Mersenne prime $M_{31} = 2^{31} - 1$ (1772, announced in a letter to Daniel Bernoulli)
• §3 Combinatorics: Latin squares + the 36 officers problem + a conjecture that was eventually overturned (1779, E530)
• §4 Number theory II: the conjectural form of quadratic reciprocity (1783, E552, left as an open problem; 13 years later the 19-year-old Gauss gave the first proof)

Each one requires a serious walk through the proof. Let us take them one section at a time.

Next section: §1 — From Newton’s two bodies to Lagrange’s five points — the relay of Euler 1751-1753 and Lagrange 1772

— Escher · §0 INTRO · v0 · 2026-05-26

§1 — The three-body problem + the Lagrange points $L_1$-$L_5$

Author of this section: Abel. Style: textbook form. Equation numbering: (1.k).

Symbol supplement for this section

In addition to the master Symbol Preamble, this section adds the following symbols:

Throughout this section normalized units are used: length scaled by $a$, mass scaled by $m_1 + m_2$, time scaled by $1/\Omega$ — so that $G(m_1 + m_2) = a^3 \Omega^2 = 1$.

1.1. The two-body problem

We start from the integrable case completely solved by Newton in the Principia (1687).

Theorem 1.1 (Kepler–Newton). Let two point masses $m_1, m_2$, under the inverse-square gravitational force

move relative to each other. Then the relative orbit $\mathbf{r}(t) := \mathbf{r}_2(t) - \mathbf{r}_1(t)$ lies in a fixed plane, and its trajectory is a conic section with the center of mass at one focus. For a bounded orbit, this conic is an ellipse, with eccentricity $e \in [0, 1)$ and semi-latus rectum $p$, parametrized as

where $\theta$ is the true anomaly measured from perihelion.

Proof. By Newton’s second law and the law of universal gravitation, the two point masses satisfy

where $\mathbf{r} := \mathbf{r}_2 - \mathbf{r}_1$, $r := |\mathbf{r}|$. Subtracting the two equations and rearranging, the relative position $\mathbf{r}$ satisfies

where the reduced mass is $\mu_r := m_1 m_2 / (m_1 + m_2)$. (1.1a) is the equation of motion of a single particle of mass $\mu_r$ in the central potential $V(r) = -G m_1 m_2 / r$.

Define the angular momentum $\mathbf{L} := \mathbf{r} \times \mu_r \dot{\mathbf{r}}$ and differentiate with respect to time:

so $\mathbf{L}$ is a constant vector. From $\mathbf{L} \perp \mathbf{r}$ and $\mathbf{L} \perp \dot{\mathbf{r}}$ we know that $\mathbf{r}$ always lies in the fixed plane through the origin with normal $\mathbf{L}$, so the orbit is a plane curve.

In this orbital plane take polar coordinates $(r, \theta)$; the velocity decomposes into the radial $\dot r$ and the tangential $r \dot\theta$, so the magnitude of the angular momentum is

Similarly one can prove conservation of energy, that is,

From (1.1b) solve for $\dot\theta = L/(\mu_r r^2)$ and substitute into the radial component of (1.1a). Change variables $u := 1/r$ and change the independent variable $t \mapsto \theta$: by the chain rule,

Into the radial equation of motion $\mu_r(\ddot r - r\dot\theta^2) = -G m_1 m_2 / r^2$ substitute $r = 1/u$, $r\dot\theta^2 = L^2 u^3/\mu_r^2$; simplifying gives (the Binet equation)

Its general solution is

where $e, \theta_0$ are constants of integration. Take $\theta_0 = 0$ (i.e., measure $\theta$ from perihelion); inverting $r = 1/u$ gives

that is, the standard form of a conic in polar coordinates, with the focus at the origin. The eccentricity $e$ determines the type of conic: $e = 0$ is a circle, $0 < e < 1$ an ellipse, $e = 1$ a parabola, $e > 1$ a hyperbola. For a bounded orbit, $E < 0$, and working back from (1.1c) gives $e \in [0, 1)$, so the orbit is an ellipse, that is, (1.2). (QED)

Hence the two-body problem is integrable in closed form: the orbit is completely determined by two constants $(p, e)$ plus the orientation angle, and given initial values it can be predicted forever.

Remark 1.1. In the formula $r = p/(1 + e\cos\theta)$, $r$ is the distance from the planet to the Sun, varying with the polar angle $\theta$: when $\theta = 0$ (perihelion), $\cos\theta = 1$, the denominator is largest, and $r$ takes its minimum; when $\theta = \pi$ (aphelion), $\cos\theta = -1$, the denominator is smallest, and $r$ takes its maximum. The eccentricity $e$ quantifies how far the ellipse departs from a circle: $e = 0$ is a perfect circle, $e \to 1$ an extremely flat ellipse. The Earth’s orbit has $e \approx 0.017$, Halley’s comet $e \approx 0.967$. $p$ controls the overall size of the ellipse.

The two-body problem is therefore integrable in closed form: given initial values, the whole orbit is completely determined by two constants $(p, e)$ plus the orientation angle, and can be predicted forever.

1.2. The non-integrability of three bodies

The reason the two-body problem can be solved is that we can find enough conserved quantities to reduce the high-dimensional phase space step by step to an equation in one variable (the Binet equation of the previous section). For three or more bodies, this procedure no longer works — that is the original meaning of “the non-integrability of the three-body problem”.

What does “non-integrable” mean? For $n$ point masses moving in three-dimensional space, the positions have $3n$ components in total and the velocities $3n$ components in total, so the phase space is $6n$-dimensional. The conserved quantities classical physics gives us number only the following $10$ in total (called the classical integrals):

In total $1 + 3 + 3 + 3 = 10$ independent conserved quantities. In the Liouville–Arnold framework of Hamiltonian mechanics (see Arnold 1989, Mathematical Methods of Classical Mechanics, §49), if a symplectic manifold has $n$ pairs of canonical coordinates and $n$ conserved quantities in involution (Poisson commuting), then the system is integrable. For two bodies ($n = 2$, phase space $12$-dimensional, i.e. $6$ canonical pairs), the $10$ quantities above plus reduction by rotational symmetry leave enough conserved quantities in involution, reducing it to a quadrature in one variable — integrable.

For three bodies ($n = 3$, phase space $18$-dimensional, i.e. $9$ canonical pairs), the $10$ quantities above, after screening for involutivity, are not enough to cover all the degrees of freedom — even after removing rotational symmetry a multidimensional problem remains, which cannot be reduced to a quadrature in one variable.

A deeper obstacle: Bruns (1887) proved that, apart from the $10$ above, the three-body problem has no additional conserved quantity of algebraic form. Poincaré (1890) further proved that there is no additional conserved quantity analytic in all the canonical coordinates. Painlevé (1898) and Sundman (1912) further pointed out that, even in certain simpler cases, the solution cannot be expressed in closed form by elementary functions, and can at most be given by an extremely slowly converging infinite series (the Sundman series, convergent but unusable in practice).

Summary in one sentence: “the non-integrability of the three-body problem” means — we cannot find enough independent conserved quantities to reduce the system of equations to something that can be solved in closed form.

Practical countermeasure — settle for second best: give up the strongest assumption, “the three bodies are dynamically equally important”. If the third mass is negligible compared with the first two — a satellite, an asteroid, a spacecraft — then its motion does not in turn affect the orbits of the first two; the first two move in ellipses (or, for simplicity, circles) as in the two-body problem, and the third only moves in the gravitational field produced by the first two. This is the restricted three-body problem. Euler studied it in 1751–53 (Theoria motus lunae), and Lagrange in 1772 (Essai sur le problème des trois corps) brought it into its modern form.

Animation of the co-rotating frame: the two primaries (orange Sun + gray Jupiter) are fixed on the $x$-axis, and the whole coordinate frame rotates about the origin with angular velocity $\Omega$. The third, test mass $m_3$ moves in the combined gravitational field produced by the two primaries + the centrifugal potential. For a static illustration see fig8 (https://timeobserver137.cyou/content/images/2026/05/fig8_rotating_frame-6.png); of the five Lagrange equilibrium points, $L_1, L_2, L_3$ are collinear, and $L_4, L_5$ form equilateral triangles with the two primaries.

1.3. The restricted three-body problem in the co-rotating frame

Setup: the two primaries $m_1, m_2$ move in circular Keplerian orbits about their common center of mass, with separation $a$ and angular velocity

Remark 1.2 (derivation of (1.4)). A direct application of Kepler’s third law to a circular two-body orbit. For a circular motion of radius $a$ by $m_1$ (relative to the center of mass), the centripetal force equals the gravitational force:

The $\dfrac{m_2}{m_1+m_2}\, a$ on the left is the distance from $m_1$ to the center of mass (the center of mass lies between $m_1, m_2$, from $m_2$ at a fraction $m_1/(m_1+m_2)$). Simplifying gives $\Omega^2 = G(m_1 + m_2)/a^3$, and taking the positive root gives (1.4).

The third mass $m_3 \ll m_1, m_2$ moves in the gravitational field of $m_1, m_2$ without perturbing their orbits. Pass to the co-rotating frame (angular velocity $\Omega$), in which $m_1, m_2$ are fixed at

where $\mu = m_2 / (m_1 + m_2) \in (0, 1/2]$ is the mass parameter, and we have normalized so that $a = 1$.

In the co-rotating frame, the equation of motion of $\mathbf{r} = (x, y, z)$ contains centrifugal and Coriolis pseudo-forces (because we have changed into a non-inertial rotating frame). Write $\rho_1 := |\mathbf{r} - \mathbf{r}_1|$ (the distance from $m_3$ to $m_1$), $\rho_2 := |\mathbf{r} - \mathbf{r}_2|$ (the distance from $m_3$ to $m_2$). The effective potential is

The three terms are respectively: the gravitational potential of $m_1$ (weight $1-\mu$, equal to $m_1/(m_1+m_2)$), the gravitational potential of $m_2$ (weight $\mu$), and the centrifugal potential induced by the co-rotating frame, $-\tfrac{1}{2}\Omega^2(x^2+y^2)$ (after normalization $\Omega = 1$).

Lemma 1.4 (Jacobi integral). In the co-rotating frame, the quantity

Sweep of the zero-velocity curves of the Jacobi integral — as $C_J$ decreases, the accessible region (green) gradually opens up from above $L_4/L_5$, through $L_1/L_2/L_3$, to around $m_1, m_2$. The core visualization of the Hill region.

is conserved along orbits of the restricted three-body problem.

Proof. The equation of motion in the co-rotating frame (including the Coriolis term $-2 \boldsymbol{\Omega} \times \dot{\mathbf{r}}$):

Take the time derivative of $C_J$, using the chain rule:

Substitute the $\ddot{\mathbf{r}}$ of (1.7a):

Expand:

The first two terms cancel; in the third term $\boldsymbol{\Omega} \times \dot{\mathbf{r}}$ is orthogonal to $\dot{\mathbf{r}}$ (a standard property of the vector cross product), so the inner product is zero. Therefore $\dot C_J = 0$, that is, $C_J$ is conserved. (QED)

The physical meaning of $C_J$: it corresponds to the “effective energy” in the co-rotating frame. In the inertial frame the total energy of the three bodies is not conserved (the restricted assumption forcibly nails $m_1, m_2$ onto circular orbits, which, compared with reality, adds an external “constraint force”), but $C_J$ is the only global conserved quantity left — it is the key tool for restricting the range in which the three-body equations can actually be analyzed to a surface of constant value.

The equilibrium points of the co-rotating motion are where the gradient of $U_{\text{eff}}$ is zero. At such a point, at rest in the co-rotating frame, taking $\dot{\mathbf{r}} = \mathbf{0}$ in (1.7a) automatically gives $\ddot{\mathbf{r}} = -\nabla U_{\text{eff}} = \mathbf{0}$, that is, the test particle stays at rest — this is the physical essence of the “Lagrange equilibrium points”.

1.4. The five equilibrium points

Theorem 1.5 (existence of the five Lagrange equilibrium points). For each $\mu \in (0, 1/2]$, the planar gradient condition $\nabla U_{\text{eff}}(x, y, 0) = \mathbf{0}$ has exactly five solutions:

Contour plot of the effective potential $U_{\text{eff}}$, showing the 5 critical points $L_1$-$L_5$ (μ = 0.05) - the three collinear points $L_1, L_2, L_3$ lie on the $x$-axis - the two equilateral points $L_4, L_5$ form equilateral triangles with $m_1, m_2$, with coordinates $(x, y) = (1/2 - \mu, \pm \sqrt{3}/2)$

Remark 1.3 (explanation of the “planar gradient condition”). The condition $\nabla U_{\text{eff}}(x, y, 0) = \mathbf{0}$ only considers equilibrium points with $z = 0$ (i.e., in the plane of $m_1, m_2$). The partial derivative with respect to $z$, $\partial_z U_{\text{eff}} = (1-\mu)z/\rho_1^3 + \mu z/\rho_2^3$, is automatically zero at $z = 0$, so $z = 0$ brings no additional constraint. All five Lagrange solutions lie in the orbital plane of the two primaries; in the region $z \neq 0$, $\partial_z U_{\text{eff}} \neq 0$, and there are no static equilibrium points.

Proof. By (1.6), the gradient in the plane is

Collinear branch ($y = 0$): equation (1.9) is automatically satisfied. Equation (1.8) becomes

Within each interval, after clearing denominators, this is a quintic equation in $x$. On the three open intervals $(-\infty, -\mu)$, $(-\mu, 1 - \mu)$, $(1 - \mu, +\infty)$, examine $f$ separately.

Differentiating (1.10) directly,

holds strictly within each interval (the gravitational terms, as inverse cubes, are negative; the centrifugal contribution is $-1$). Hence $f$ is strictly decreasing on each interval. The limits at the endpoints are respectively: - $(-\infty, -\mu)$: $f(x) \to +\infty$ as $x \to -\infty$ (the centrifugal $-x$ dominates and is positive), $f(x) \to -\infty$ as $x \to -\mu^-$ (the gravity from $m_1$ diverges, negative). - $(-\mu, 1-\mu)$: $f(x) \to +\infty$ as $x \to -\mu^+$ (the sign of gravity flips), $f(x) \to -\infty$ as $x \to (1-\mu)^-$. - $(1-\mu, +\infty)$: $f(x) \to +\infty$ as $x \to (1-\mu)^+$, $f(x) \to -\infty$ as $x \to +\infty$.

By strict monotonicity and the intermediate value theorem, $f$ has exactly one root in each interval. Write them $L_3 < -\mu$, $-\mu < L_1 < 1-\mu$, $L_2 > 1-\mu$ (modern convention: $L_1$ between the two primaries, $L_2$ beyond the smaller mass, $L_3$ opposite the smaller mass).

Animation of the order in which the three collinear Lagrange points appear: $L_1$ between the Sun and Jupiter, $L_2$ beyond Jupiter, $L_3$ opposite the Sun. Each point corresponds to the unique zero of equation (1.10) in a different interval.

Triangular branch ($y \neq 0$): the square bracket in (1.9) must be zero, that is,

Combined with the symmetric constraint that (1.8) imposes on the two gravitational terms, the symmetric solution is $\rho_1 = \rho_2 = 1$ — that is, $m_3$ is at equal distances from the two primaries, exactly equal to the separation of the two primaries. Substituting into (1.10b) gives $(1-\mu) + \mu = 1$ ✓; substituting into (1.8) gives $(1-\mu)(x+\mu) + \mu(x-1+\mu) - x = x - x = 0$ ✓. The two solutions are

Each of them forms an equilateral triangle with $m_1, m_2$. As for the argument that this is the unique triangular solution — (1.10b) together with (1.8) gives a system of equations in $(\rho_1, \rho_2)$, which over the positive reals can be proved to have only the one pair of roots $(1, 1)$ — for the detailed proof see Szebehely 1967, Theory of Orbits, §4.4. (QED)

1.5. Linear stability — the Routh criterion

In the Sun-Jupiter co-rotating frame, the libration orbits of 30 Trojan test particles near $L_4$ (μ = 0.001, t ∈ [0, 200] rotating frame units). Because μ < μ_R ≈ 0.0385, the particles undergo bounded libration near $L_4$ rather than escaping — a dynamical demonstration of Routh stability.

§1.4 has found five equilibrium points. However, the mere existence of an equilibrium is not enough to guarantee that a spacecraft or celestial body can stay there — the tip of a pencil standing upside down is a mathematical equilibrium point, but any perturbation destabilizes it. So we must further examine how each equilibrium point responds to small perturbations: if the perturbation decays over time or oscillates boundedly, it is called stable; if it diverges exponentially, unstable.

Let $\mathbf{r} = \mathbf{r}_* + \boldsymbol{\xi}$, where $\boldsymbol{\xi}$ is a small perturbation. Substitute into the co-rotating equation of motion (1.7a) and Taylor-expand $\nabla U_{\text{eff}}$ at $\mathbf{r}_*$ (the zeroth-order term vanishes, since $\mathbf{r}_*$ is an equilibrium point), keeping up to linear terms, to get

where $H := \big(\partial_i \partial_j U_{\text{eff}}\big)_{i,j \in \{x,y\}}$ is the Hessian matrix of $U_{\text{eff}}$ restricted to the $xy$ plane, evaluated at $\mathbf{r}_*$. Since (1.12) is a system of linear second-order ordinary differential equations with constant coefficients, substituting the exponential trial solution $\boldsymbol{\xi} \sim e^{\lambda t}$ gives the characteristic equation

where $U_{xx}, U_{yy}, U_{xy}$ are evaluated at $\mathbf{r}_*$. (1.13) is a biquadratic equation in $\lambda^2$, with $4$ roots $\lambda_k$ in all; the equilibrium point is stable against linear perturbations if and only if all $\lambda_k$ have non-positive real parts (the perturbation oscillates or decays, and does not diverge exponentially). By the biquadratic structure of (1.13), this condition is equivalent to both roots in $\lambda^2$ being negative real numbers.

Collinear $L_1, L_2, L_3$: direct computation (Charlier 1899) gives $U_{xx} < 0$ and $U_{yy} > 0$ at each collinear equilibrium point, so the determinant $U_{xx} U_{yy} - U_{xy}^2 < 0$. Equation (1.13) is biquadratic in $\lambda^2$ with a negative constant term; the two values of $\lambda^2$ have opposite signs, so there is a positive real $\lambda$. The collinear Lagrange points are linearly unstable for all $\mu \in (0, 1/2]$.

Equilateral $L_4, L_5$: we compute the components of the Hessian at $L_4 = (1/2 - \mu, \sqrt 3/2, 0)$. Taking second derivatives of (1.6):

Substitute the coordinates of $L_4$, $(x, y) = (1/2 - \mu, \sqrt 3/2)$, and the equilateral condition $\rho_1 = \rho_2 = 1$; noting $x + \mu = 1/2$, $x - 1 + \mu = -1/2$, $y^2 = 3/4$, standard simplification gives

where the sign of $U_{xy}$, at $L_4$, is taken as $-$, and at $L_5$ as $+$ (because at $L_5$, $y = -\sqrt 3/2$). Hence

(The key algebraic identity $1 - (1 - 2\mu)^2 = 4\mu(1-\mu)$ — a direct difference-of-squares expansion.) From $\Omega = 1$ (normalization), in (1.13) $4\Omega^2 = 4$, so

Let $z := \lambda^2$; the quadratic equation $z^2 + z + \tfrac{27}{4}\mu(1-\mu) = 0$ has solutions

For $\lambda$ to be purely imaginary (i.e., the perturbation oscillates and does not grow), $z$ must take two negative real roots, so (a) the discriminant $1 - 27\mu(1-\mu) \geq 0$ (otherwise $z$ is complex and $\lambda$ has a real part); (b) both roots $z_\pm$ are negative (by Vieta: $z_+ z_- = \tfrac{27}{4}\mu(1-\mu) > 0$, $z_+ + z_- = -1 < 0$, automatically negative). Hence stable ⟺

Theorem 1.6 (Routh; Adams Prize essay 1872, monograph On the Stability of Motion 1877). The equilateral Lagrange points $L_4, L_5$ are linearly stable if and only if $\mu < \mu_{\text{R}} \approx 0.0385$, where $\mu_{\text{R}} = \frac{1}{2}(1 - \sqrt{23/27})$.

Visualization of the Routh stability criterion: the position of the curve $\mu(1-\mu)$ relative to the threshold $1/27$. The blue region is stable ($L_4, L_5$ linearly stable), the red region unstable. Three sample systems are marked — Sun-Earth ($\mu \approx 3 \times 10^{-6}$, deeply stable), Sun-Jupiter ($\mu \approx 9.5 \times 10^{-4}$, stable), Pluto-Charon ($\mu \approx 0.117$, unstable, cannot sustain Trojans).

1.6. Worked example — the distance of the Sun-Earth $L_1$

Take $m_1$ = the Sun (symbol $\odot$), $m_2$ = the Earth (symbol $\oplus$). By the definition in §1.3, the mass parameter $\mu = m_2/(m_1 + m_2)$ is here written concretely as $\mu_{\oplus}$ to indicate that it is the “Sun-Earth coefficient”:

(Since $M_{\oplus} \ll M_{\odot}$, the denominator is approximately $M_{\odot}$, so $\mu_{\oplus} \approx M_{\oplus}/M_{\odot}$.)

The collinear equilibrium point $L_1$ between the Sun and the Earth satisfies (1.10) on the interval $(-\mu, 1 - \mu)$. Let $\xi := 1 - \mu - x$ be the distance of $L_1$ from the Earth toward the Sun (normalized, one unit = $1\,\text{AU}$); substitute $x = 1 - \mu - \xi$ into (1.10); on the interval $(-\mu, 1-\mu)$, $x + \mu > 0$ and $x - 1 + \mu < 0$, so $|x+\mu| = 1 - \xi, |x-1+\mu| = \xi$, and the equation becomes

To find the dominant balance, when $\mu \ll 1$ and $\xi \ll 1$ the terms Taylor-expand (to first order) as follows: - $\dfrac{1-\mu}{(1-\xi)^2} \approx (1-\mu)(1+2\xi) \approx 1 + 2\xi - \mu - 2\mu\xi \approx 1 + 2\xi - \mu$. - $-\dfrac{\mu}{\xi^2}$ — the dominant term in the competition between the relative magnitudes of $\mu$ and $\xi$. - $-(1-\xi-\mu) = -1 + \xi + \mu$.

Combining the three terms, (1.18) becomes

Here $1$ and $-1$ cancel, $-\mu$ and $+\mu$ cancel, leaving the gravity-gradient term $2\xi$ (from the Taylor expansion of $1/(1-\xi)^2$) and the centrifugal position term $+\xi$ (from, within $-(1-\xi-\mu)$, the $+\xi$), which combine into $3\xi$. The dominant balance is $3\xi \approx \mu/\xi^2$, that is,

(The factor $3 =$ gravity-gradient contribution $2$ + centrifugal position contribution $1$.) Substituting the numbers,

The Earth is $1\,\text{AU} = 1.496 \times 10^8\,\text{km}$ from the Sun, so $L_1$ is located at approximately

that is, about 1.5 million kilometers from the Earth in the direction of the Sun, matching exactly the halo orbit of the SOHO mission near that point (1995 to the present).

For $L_2$, a similar expansion on $(1 - \mu, \infty)$ gives $\xi_{L_2} \approx (\mu/3)^{1/3}$ (on the side away from the Sun) — the station-keeping orbit of JWST since 2022.

For the Sun-Jupiter system, $\mu_{\textrm{J}} \approx 9.5 \times 10^{-4}$, which satisfies $\mu_{\textrm{J}} < \mu_{\text{R}}$, so $L_4, L_5$ are linearly stable — and in fact each of them hosts a considerable number of Trojan asteroids: more than $12{,}000$ have been numbered and recorded (including named ones), while extrapolation from the IRAS survey estimates the total number of Jupiter Trojans larger than 1 km at more than $10^6$ (the “Greek camp” at $L_4$ and the “Trojan camp” at $L_5$, named starting with Wolf’s first one, 588 Achilles, in 1906).

1.7. Modern significance

The Lagrange equilibrium points are far from an academic curiosity; they organize a whole set of programs of contemporary space science.

• JWST (NASA/ESA/CSA, launched 2021-12-25) stays in a halo orbit around the Sun-Earth $L_2$, using the shading on the side away from the Sun to maintain the thermal stability of its 6.5-m primary mirror.
• SOHO (NASA/ESA, 1995 to the present) operates in a halo orbit at the Sun-Earth $L_1$, continuously observing the solar disk.
• DSCOVR (NOAA, 2015 to the present) and the soon-to-launch ARIEL (ESA, 2029) likewise take $L_1$ or $L_2$ as their target.
• The STEREO-A/B twin probes (NASA, 2006 to the present) patrol the trailing and leading regions near the Sun-Earth $L_4, L_5$.
• The Hilda asteroid family and the Trojan asteroid swarms at the Sun-Jupiter $L_4, L_5$ provide a natural laboratory for testing the $\mu_{\text{R}}$ criterion on geological time scales.

Routh’s criterion $\mu < 1/27$ (1.16) explains why we can see Trojans at Jupiter, Mars, Neptune, and even the Earth — the two Earth Trojans confirmed so far, the asteroids 2010 TK$_7$ (confirmed 2011) and 2020 XL$_5$ (confirmed 2021), both lie at the Sun-Earth $L_4$ — while the Pluto–Charon system ($\mu \approx 0.117$, beyond the Routh threshold) cannot dynamically sustain Trojans, and so none have ever been observed.

Stability comparison of the $L_1$ saddle point (red) vs the $L_4$ equilibrium (green). The initial perturbations are of the same magnitude (position < 0.005 / velocity < 0.002 units), but the particles near $L_1$ escape within about 30 rotating-frame time units, while the particles near $L_4$ keep librating — a direct visualization of saddle-point instability vs equilateral stability.

The quintic equation Euler wrote down in 1751, refined by Lagrange in 1772 and with its stability determination completed by Routh in 1875, controls in 2026 the orbit of a $1-billion telescope — one of the clearest examples in mathematical physics of formal analysis, acting on a superficially simple equation, accumulating and releasing its power.

— End of §1.

§2 — The Mersenne prime $M_{31}$

Author of this section: Abel. Style: textbook form. Equation numbering: (2.k).

Symbol supplement for this section

This section works over $\mathbb{Z}$ and over finite fields $\mathbb{F}_q$ ($q$ prime). In addition to the master Symbol Preamble, it adds:

Unless stated otherwise, $p, q$ both denote odd primes.

2.1. Definition and the 1644 list

Definition 2.1. For a positive integer $p$, the $p$-th Mersenne number is defined as

If $M_p$ is prime, it is called a Mersenne prime.

Proposition 2.2 (Mersenne’s first reduction). If $M_p$ is prime, then $p$ must be prime.

Proof. Suppose to the contrary that $p = ab$ ($1 < a, b$). Let $X := 2^a$; then $M_p = 2^{ab} - 1 = X^b - 1$. Use the algebraic identity (sum of a geometric series)

(substituting $X = 1$: right side $(1-1)\cdot b = 0$ ✓; $X = 2$: $(2^b-1) = 1 \cdot (2^{b-1} + \cdots + 1) = 2^b - 1$ ✓). Substituting back $X = 2^a$:

From $1 < a < p$, $2^a - 1$ is an integer greater than $1$; also $2^a - 1 < 2^{ab} - 1$ (because $a < ab$), so $2^a - 1$ is a nontrivial factor of $M_p$. $M_p$ is composite. (QED)

(Note: (2.2) is not the binomial expansion $(1+1)^n$, but the geometric-series factorization of the polynomial $X^b - 1$. This is a standard algebraic identity, valid over every commutative ring.)

The converse does not hold — many primes $p$ give composite $M_p$. In 1644 (Cogitata Physico-Mathematica) Marin Mersenne listed, as those for which he claimed $M_p$ is prime, the $p \leq 257$:

Mersenne’s list contains five errors, discovered one after another between 1903 and 1947: - $M_{67}$ and $M_{257}$ are in fact composite (Cole 1903 gave the factorization of $M_{67}$; Kraitchik 1922 and Lehmer 1947 gave $M_{257}$). - $M_{61}, M_{89}, M_{107}$ are actually prime but were left out (Pervushin 1883 / Powers 1911 / Powers 1914).

The first nontrivial entry that needs verification by hand is $M_{31} = 2{,}147{,}483{,}647$ — a ten-digit number, a rather formidable task by trial division. In 1772, completely blind, Euler completed the verification by mental arithmetic and sent the result by letter from Petersburg to Basel to inform Daniel Bernoulli. The argument relies on three layers of sieves that progressively shrink the trial-division space.

2.2. Sieve 1 — Fermat’s little theorem

Lemma 2.3 (Fermat’s little theorem). For any prime $q$ and, with $\gcd(a, q) = 1$, any integer $a$,