on the proof of Sendov's conjecture

Overview

Sendov’s conjecture asserts that every zero of a polynomial whose zeroes lie in the closed unit disk is within distance one of a critical point. Lech Mazur announced a computer-assisted proof on August 5, 2026; Terence Tao subsequently extracted a substantially shorter, human-readable argument (Mazur, 2026; Tao, 2026).

The conceptual core is elementary. In Tao’s streamlined route, normalize a distinguished zero to $a\in [0,1]$ and write every putative distant critical point as $a-1/q_j$, with $|q_j|\le 1$. Four identities obtained from coefficients and from evaluating $p$ or $p^{\prime }$ at $0$, $a$, and the reflected point $1/a$ are the only remaining communication between the other zeroes $z_j$ and the reciprocal critical data $q_j$. A polar identity forces the mean of the $q_j$ close to $1$; two origin identities, Maclaurin’s inequality, and a finite-product defect estimate force it away from the same region. The problem thereby collapses to incompatible inequalities for two real scalars.

The reduction, the boundary case, and degrees $2$ through $5$ are hand proofs. One qualification is essential: a bounded range of the final one-variable inequality is closed by exact computer verification. Tao’s companion development implements this check with rational Bernstein certificates and treats the remaining degrees analytically (Tao, 2026b). This note explains the certificate’s mathematical content but does not reproduce its thousands of rational coefficients. No Lean source file was inspected, and no Lean code was run for this note.

🏷️ The Conjecture and Its Strict Form

Write

$$\overline{\mathbb{D}}=\{z\in \mathbb{C}:|z|\le 1\}.$$

Sendov's conjecture

Let $p$ be a complex polynomial of degree $n\ge 2$, all of whose zeroes lie in $\overline{\mathbb{D}}$. For every zero $a$ of $p$, there is a zero $\zeta$ of $p^{\prime }$ such that

$$|\zeta -a|\le 1.$$

The stronger Phelps—Rodriguez statement classifies equality (Phelps & Rodriguez, 1972).

Phelps--Rodriguez theorem

Under the same hypotheses, one may take $|\zeta -a|<1$ unless $|a|=1$ and

$$p(z)=c(z^n-a^n)$$

for some $c\ne 0$.

The statement that drives the proof is the following strict interior form.

Strict interior theorem

Under the hypotheses of Sendov’s conjecture, let $a$ be a zero with $|a|<1$. Then $p^{\prime }$ has a zero $w$ satisfying

$$|w-a|<1.$$

Once this is known, only the case $|a|=1$ remains. The boundary argument below either produces a critical point at distance strictly less than one or forces the extremal polynomial $c(z^n-a^n)$. Thus the strict interior theorem and the boundary classification together imply the Phelps—Rodriguez statement, which in turn implies Sendov’s conjecture.

The problem was recorded in Hayman’s collection of research problems (Hayman, 1967). Brown and Xiang proved it through degree eight (Brown & Xiang, 1999), while Tao proved it for every sufficiently large degree (Tao, 2022). Those results left a finite but ineffective middle range. Mazur’s proof closed every degree, and Tao’s digestion also proves the strict interior statement, hence the full Phelps—Rodriguez theorem.

Status and proof boundary

These sources were only days old when this note was written. Mazur’s manuscript explicitly separates its written analytic reduction, its exact supplementary certificates, and a Lean formalization that proves the theorem by a streamlined route rather than checking the manuscript line by line (Mazur, 2026). Tao’s second formalization follows Tao’s streamlined proof and shares no code with the first, but it is not an independent mathematical proof (Tao, 2026b). Formal verification is strong evidence for the theorem; it is not a substitute for independent specialist review of the exposition and provenance.

Mazur’s manuscript and Tao’s digestion should also be distinguished mathematically. Mazur begins with a minimum-distance scaling and reaches a scalar inequality in parameters customarily denoted $m$ and $\lambda$; its finite certificate covers $5\le m\le 499$ and a second exact certificate enters the analytic tail. Tao reorganizes the same proof lineage around four communication identities and the two parameters $\alpha$ and $\beta (1)$. The argument developed below is Tao’s streamlined descendant, with Mazur’s manuscript used as an independent check on the normalization, defect estimate, low-degree argument, and certification boundary.

🏷️ Preliminary Geometry and Inequalities

All zeroes and critical points in this note are counted with algebraic multiplicity. The fundamental theorem of algebra gives

$$p(z)=c\prod _{k=1}^n(z-{\lambda }_k),p^{\prime }(z)=cn\prod _{j=1}^{n-1}(z-w_j).$$

A zero $a$ of multiplicity at least two is automatically a critical point, so Sendov’s conclusion is immediate at such a zero. Any hypothetical counterexample therefore has a simple distinguished zero.

Gauss--Lucas as background

The Gauss—Lucas theorem places every critical point in the convex hull of the zeroes, hence in $\overline{\mathbb{D}}$ here. A short proof illustrates why Sendov is a genuinely sharper, local assertion. Away from the zeroes,

$$\frac{p^{\prime }(z)}{p(z)}=\sum _{k=1}^n\frac{1}{z-{\lambda }_k}.$$

If $z$ lies outside the convex hull, a rotation makes every $z-{\lambda }_k$ have positive real part. Then every reciprocal $(z-{\lambda }_k{)}^{-1}$ also has positive real part, so their sum cannot vanish. Gauss—Lucas gives the global disk containing the critical points; Sendov asks for a unit disk centered at each individual zero.

Multiplying $p$ by a nonzero constant changes neither its zeroes nor its critical points. Rotation is equally harmless: if $|\omega |=1$, then $P(z)=p(\omega z)$ has zeroes and critical points obtained from those of $p$ by multiplication by $\overline{\omega }$, and all relevant distances are unchanged. Consequently a nonzero distinguished zero can be rotated to a real number $a\in (0,1]$.

For $0\le a<1$ and $|z|\le 1$, the elementary reflection identity

$$|1-az{|}^2-|a-z{|}^2=(1-a^2)(1-|z{|}^2)\ge 0$$

is equivalent to the fact that $(a-z)/(1-az)$ maps the closed unit disk to itself. This is the sole Möbius-geometric input in the interior proof.

Two inequalities compress products of nonnegative numbers. For $r_1,\dots ,r_m\ge 0$, quadratic-mean—geometric-mean gives

$$\prod _{j=1}^m{r}_j\le {\left(\frac{1}{m}\sum _{j=1}^m{r}_j^2\right)}^{m/2},$$

while the codimension-one case of Maclaurin’s inequality gives

$$\frac{1}{m}\sum _{j=1}^m\prod _{k\ne j}{r}_k\le {\left(\frac{1}{m}\sum _{j=1}^m{r}_j\right)}^{m-1}.$$

Proof: The required Maclaurin inequality by smoothing

The cases $m=1,2$ are immediate. For $m\ge 3$, fix the sum of the $r_j$. If two variables $u,v$ are replaced by their average, then the codimension-one elementary symmetric polynomial changes only through a term

$$uv{e}_{m-3}(\widehat{u,v})+(u+v)e_{m-2}(\widehat{u,v}),$$

where the second summand is fixed and the first cannot decrease because $uv\le ((u+v)/2{)}^2$. Repeated averaging therefore maximizes the expression when every $r_j$ equals their common mean. At that point the left side is exactly the right side. Zero variables follow by continuity.

We will also use, with the signs indicated,

$$1+u\le e^u(u\in \mathbb{R}),$$

and

$$|s+it|\le s+\frac{t^2}{2s}(s>0).$$

The latter follows after squaring, since

$${\left(s+\frac{t^2}{2s}\right)}^2-(s^2+t^2)=\frac{t^4}{4s^2}\ge 0.$$

Two model cases

For a quadratic with zeroes $a$ and $z$, the unique critical point is $(a+z)/2$, so its distance from $a$ is $|z-a|/2\le 1$. At the opposite extreme, if $|a|=1$ and $p(z)=z^n-a^n$, then the only critical point is $0$, of multiplicity $n-1$, and $|0-a|=1$. This is exactly the equality family in Phelps—Rodriguez.

🏷️ A Counterexample in Reciprocal Coordinates

Assume that the strict interior theorem fails. Rotate the plane and multiply by a scalar so that the distinguished zero is real and $p$ is monic:

$$0\le a<1,p(a)=0.$$

Every critical point $w_j$ then satisfies $|w_j-a|\ge 1$. Put $m=n-1$ and introduce the reciprocal displacement

$$w_j=a-\frac{1}{q_j},q_j=\frac{1}{a-w_j},0<|q_j|\le 1,1\le j\le m.$$

The distinguished zero is simple, since otherwise $p^{\prime }(a)=0$ and the strict conclusion is already true. Let $z_1,\dots ,z_m$ be the remaining zeroes of $p$, counted with multiplicity, so $|z_j|\le 1$ and $z_j\ne a$. Thus

$$p(z)=(z-a)\prod _{j=1}^m(z-z_j)$$

and

$$p^{\prime }(z)=n\prod _{j=1}^m\left(z-a+\frac{1}{q_j}\right).$$

The fundamental theorem of calculus along the segment $a+t(z-a)$ gives

$$p(z)-p(a)=(z-a){\int }_0^1{p}^{\prime }(a+t(z-a))dt.$$

Since $p(a)=0$, substitution of the critical-point factorization gives the antiderivative identity

$$p(z)=(z-a){\int }_0^{1}n\prod _{j=1}^m\left(t(z-a)+\frac{1}{q_j}\right)dt.$$

The case $a=0$ already contradicts these factorizations. Indeed, differentiating $p(z)=z{\prod }_j(z-z_j)$ at the origin and evaluating the critical-point factorization there give

$$p^{\prime }(0)=\prod _{j=1}^m(-z_j)=n\prod _{j=1}^m\frac{1}{q_j}.$$

The left side has modulus at most $1$, while the right side has modulus

$$n\prod _{j=1}^m|q_j{|}^{-1}\ge n\ge 2.$$

Hence the substantive interior case has

$$0<a<1.$$

🏷️ Four Communication Identities

Define

$$F(t)=\prod _{j=1}^m(1-at{q}_j).$$

The two point clouds $(z_j)$ and $(q_j)$ are connected by the following identities. The apparently singular expression in the last one is always understood in the polynomial, division-free form

$$\left(\prod _{j=1}^m{z}_j\right)\left(1+a\sum _{j=1}^m\frac{1}{z_j}\right):=\prod _{j=1}^m{z}_j+a\sum _{j=1}^m\prod _{k\ne j}{z}_k.$$

It therefore remains meaningful when some $z_j$ vanishes.

Communication identities

The centroid identity is

$$\frac{1}{n}\left(a+\sum _{j=1}^m{z}_j\right)=\frac{1}{m}\sum _{j=1}^m\left(a-\frac{1}{q_j}\right).$$

The polar identity is

$$\prod _{j=1}^m\frac{1-a{z}_j}{a-z_j}={\int }_0^1\prod _{j=1}^m(a+t(1-a^2)q_j)dt.$$

The first and second origin identities are

$$(-1{)}^m\prod _{j=1}^m{z}_j=\frac{n}{{\prod }_{j=1}^m{q}_j}{\int }_0^{1}F(t)dt$$

and

$$(-1{)}^m\prod _{j=1}^m{z}_j\left(1+a\sum _{j=1}^m\frac{1}{z_j}\right)=\frac{n}{{\prod }_{j=1}^m{q}_j}F(1).$$

Proof: Coefficients and three evaluations

Put $S_z=a+{\sum }_j{z}_j$ and $S_w={\sum }_j{w}_j$. Since $m=n-1$, the coefficient of $z^{m-1}$ in the derivative of

$$p(z)=z^n-S_z{z}^m+\text{terms of degree at most }m-1$$

is $-m{S}_z$. On the other hand,

$$p^{\prime }(z)=n\prod _{j=1}^m(z-w_j)$$

has coefficient $-n{S}_w$ at $z^{m-1}$. Thus $m{S}_z=n{S}_w$, which is precisely the centroid identity. This coefficient calculation remains literal when $m=1$.

For the polar identity, put $\delta =1-a^2>0$ and compute $p(1/a)/p^{\prime }(a)$ twice. The simplicity of $a$ gives

$$p^{\prime }(a)=\prod _{j=1}^m(a-z_j)=n\prod _{j=1}^m\frac{1}{q_j}\ne 0.$$

The zero factorization therefore yields

$$\begin{array}{rl}\frac{p(1/a)}{p^{\prime }(a)}& =\frac{(1/a-a){\prod }_j(1/a-z_j)}{{\prod }_j(a-z_j)}\\ & =\frac{\delta }{a^{m+1}}\prod _{j=1}^m\frac{1-a{z}_j}{a-z_j}.\end{array}$$

The antiderivative representation yields instead

$$\begin{array}{rl}\frac{p(1/a)}{p^{\prime }(a)}& =\frac{\delta }{a}{\int }_0^1\prod _{j=1}^m\left(1+\frac{t\delta q_j}{a}\right)dt\\ & =\frac{\delta }{a^{m+1}}{\int }_0^1\prod _{j=1}^m(a+t\delta q_j)dt.\end{array}$$

Cancelling the common positive factor $\delta /a^{m+1}$ gives the polar identity.

Evaluating $p(0)$ in the zero and antiderivative factorizations gives

$$-a(-1{)}^m\prod _j{z}_j=-an{\int }_0^1\prod _j\left(-at+\frac{1}{q_j}\right)dt,$$

which is the first origin identity after cancelling $-a$ and extracting ${\prod }_j{q}_j^{-1}$. Finally, direct differentiation of the zero factorization gives

$$p^{\prime }(0)=(-1{)}^m\left(\prod _j{z}_j+a\sum _j\prod _{k\ne j}{z}_k\right),$$

whereas the critical-point factorization gives

$$p^{\prime }(0)=\frac{n}{{\prod }_j{q}_j}\prod _j(1-a{q}_j)=\frac{n}{{\prod }_j{q}_j}F(1).$$

Equality of these two division-free expressions is the second origin identity.

No further property of $p$ is used. From this point onward, one studies two multisets in the closed disk satisfying these four relations. This deliberate loss of information is the decisive simplification: the zeroes and critical points cease to be a tightly coupled algebraic system and communicate only through a few scalar statistics.

🏷️ The Two Scalar Parameters

Let

$$\mu =x+iy:=\frac{1}{m}\sum _{j=1}^m{q}_j.$$

Since the closed disk is convex,

$$x^2+y^2\le 1,|y|\le \sqrt{1-x^2}.$$

Two further quantities encode the boundary scale of $a$ and the displacement of the mean:

$$\alpha :=\frac{m}{2}(1-a^2)$$

so that

$$0<\alpha <\frac{m}{2},a^2=1-\frac{2\alpha }{m}.$$

and

$$\beta (t):=1-2axt+a^2{t}^2.$$

Write $b:=\beta (1)$. Then

$$b=1-2ax+a^2=(1-a{)}^2+2a(1-x)=1-x^2+(x-a{)}^2.$$

Thus $\alpha =O(1)$ is the delicate boundary layer $a=1-O(1/n)$, while $b$ measures how close the real part of the critical-point mean is to its extremal value. In particular,

$$b>0,1-x^2\le b.$$

The proof produces an upper constraint on $b$ from the polar identity and a competing lower constraint from the origin identities.

🏷️ The Polar Channel

For real $0<a<1$, the reflection identity from the preliminaries gives

$$|a-z_j|\le |1-a{z}_j|.$$

The denominator $a-z_j$ is nonzero because the distinguished zero is simple. Hence

$$\left|\frac{1-a{z}_j}{a-z_j}\right|\ge 1.$$

The polar identity and the triangle inequality therefore imply

$$1\le {\int }_0^1\prod _{j=1}^m\left|a+t(1-a^2)q_j\right|dt.$$

For every $t\in [0,1]$, quadratic-mean—geometric-mean and $|q_j|\le 1$ give

$$\begin{array}{rl}\prod _{j=1}^m\left|a+t(1-a^2)q_j\right|& \le {\left(\frac{1}{m}\sum _{j=1}^m{\left|a+t(1-a^2)q_j\right|}^2\right)}^{m/2}\\ & \le {\left(a^2+2atx(1-a^2)+t^2(1-a^2{)}^2\right)}^{m/2}.\end{array}$$

Consequently,

$$1\le {\int }_0^1{\left(a^2+2atx(1-a^2)+t^2(1-a^2{)}^2\right)}^{m/2}dt.$$

Put $\delta =1-a^2=2\alpha /m$. Since $t^2\le t$ and

$$2ax+(1-a^2)=2-b,$$

the expression inside the power satisfies

$$1+\frac{2\alpha }{m}(-1+(2-b)t).$$

Indeed, this follows directly from

$$\begin{array}{rl}{a}^2+2atx\delta +t^2{\delta }^2& =1+\delta (-1+2axt+\delta t^2)\\ & \le 1+\delta (-1+(2ax+\delta )t),\end{array}$$

and $2ax+\delta =2-b$. Raising to the power $m/2$ and using $1+u\le e^u$ gives the exponential polar inequality

$$1<{\int }_0^1\exp \left(\alpha (-1+(2-b)t)\right)dt.$$

The inequality is strict after integration: for $0\le t<1$, either $t^2<t$ makes the preceding relaxation strict or $1+u<e^u$ does, apart from at most isolated values of $t$.

First, $b<1$. Otherwise the affine exponent $\alpha (-1+(2-b)t)$ is nonpositive on $[0,1]$ and negative away from at most the endpoint, making the integral strictly less than one. Since $2-b>0$, evaluation now gives

$$1<\frac{e^{\alpha (1-b)}-e^{-\alpha }}{\alpha (2-b)}.$$

Equivalently,

$$e^{\alpha (1-b)}>\alpha (2-b)+e^{-\alpha }\ge \alpha .$$

Because $2-b>1$, the displayed lower bound is strictly stronger than what is needed below; in particular,

$$b<1-\frac{\log \alpha }{\alpha }.$$

A second elementary estimate is stronger for small $\alpha$.

Hyperbolic-sine estimate

For $h>0$,

$$\log \frac{\sinh h}{h}\le \sqrt{h^2+9}-3.$$

Both sides vanish at $h=0$. Comparing derivatives reduces the claim to

$$h^4{\sinh }^{2}h-(h^2+9)(h\cosh h-\sinh h{)}^2\ge 0.$$

Its Taylor expansion is

$$\sum _{k=4}^{\infty }\frac{2^{2k-3}(2k-1)(2k-6{)}^2}{(2k)!}{h}^{2k},$$

whose coefficients are nonnegative.

Indeed, with

$$u=\frac{\alpha b}{2},h=\frac{\alpha (2-b)}{2},$$

the exponential integral equals

$$e^{-u}\frac{\sinh h}{h}.$$

Its being larger than $1$ gives

$$u<\log \frac{\sinh h}{h}\le \sqrt{h^2+9}-3.$$

Since $h=\alpha -u$, both sides of $u+3<\sqrt{h^2+9}$ are positive. Squaring and cancelling $u^2+9$ gives

$$2u(\alpha +3)<{\alpha }^2.$$

Substituting $u=\alpha b/2$ yields

$$b<\frac{\alpha }{3+\alpha }.$$

We have proved the polar constraint

$$0<b<\min \left\{\frac{\alpha }{3+\alpha },1-\frac{\log \alpha }{\alpha }\right\}.$$

Since $1-b=2a(x-a/2)$, it also follows that

$$x>\frac{a}{2}>0.$$

🏷️ The Origin Channel

Differentiate $F(t)={\prod }_j(1-at{q}_j)$. Adding and subtracting the mean term gives the exact identity

$$F^{\prime }(t)=-ma\mu F(t)-a^{2}t\sum _{j=1}^m{q}_j^2\prod _{k\ne j}(1-at{q}_k).$$

To check the identity, begin with direct product differentiation and use

$$q_j\prod _{k\ne j}(1-at{q}_k)=q_{j}F(t)+at{q}_j^2\prod _{k\ne j}(1-at{q}_k),$$

then sum over $j$ and recall that ${\sum }_j{q}_j=m\mu$.

Let $r_j=|1-at{q}_j|$. Using $|q_j{|}^2\le 1$, followed by the top Maclaurin inequality and Cauchy—Schwarz, gives

$$\begin{array}{rl}\sum _{j=1}^m\prod _{k\ne j}{r}_k& \le m{\left(\frac{1}{m}\sum _{j=1}^m{r}_j\right)}^{m-1}\\ & \le m{\left(\frac{1}{m}\sum _{j=1}^m{r}_j^2\right)}^{(m-1)/2}\\ & \le m\beta (t{)}^{(m-1)/2}.\end{array}$$

The last step uses the exact mean-square calculation

$$\begin{array}{rl}\frac{1}{m}\sum _{j=1}^m{r}_j^2& =1-2axt+a^2{t}^2\frac{1}{m}\sum _{j=1}^m|q_j{|}^2\\ & \le 1-2axt+a^2{t}^2=\beta (t).\end{array}$$

It follows that

$$\left|F^{\prime }(t)+ma\mu F(t)\right|\le a^{2}tm\beta (t{)}^{(m-1)/2}.$$

Since $F(0)=1$, integration yields

$$1\le \left|F(1)+ma\mu {\int }_0^{1}F(t)dt\right|+a^{2}m{\int }_0^{1}t\beta (t{)}^{(m-1)/2}dt.$$

The origin identities convert the first term into an expression involving the product

$$J:=\prod _{j=1}^m{z}_j{q}_j,|J|\le 1.$$

The only obstacle is that inversion agrees with conjugation exactly only on the unit circle. The following defect estimate measures the error inside the disk.

Finite-product defect lemma

If $w_1,\dots ,w_N\in \overline{\mathbb{D}}$, then

$$\left(\prod _{j=1}^N|w_j|\right)\sum _{j=1}^N\left|\frac{1}{w_j}-\overline{w_j}\right|\le 1-\prod _{j=1}^N|w_j{|}^2.$$

When a factor vanishes, the left side is interpreted in the equivalent division-free form

$$\sum _{j=1}^N(1-|w_j{|}^2)\prod _{k\ne j}|w_k|.$$

Proof: Radial defects are superadditive

The division-free form shows that zero factors follow by continuity, so assume every $w_j\ne 0$ and write $|w_j|=e^{-s_j}$ with $s_j\ge 0$. Then

$$\left|\frac{1}{w_j}-\overline{w_j}\right|=2\sinh s_j$$

and

$$1-\prod _j|w_j{|}^2=2\left(\prod _j|w_j|\right)\sinh \left(\sum _j{s}_j\right).$$

The result is therefore the superadditivity

$$\sinh \left(\sum _j{s}_j\right)\ge \sum _j\sinh s_j,$$

which follows by induction from the addition formula for $\sinh$.

Apply this lemma to the combined family

$$z_1,\dots ,z_m,\overline{q_1},\dots ,\overline{q_m}.$$

Here is the substitution in detail. The conjugate of the centroid identity rearranges to

$$J\sum _{j=1}^m\left(\frac{m}{n}\overline{z_j}+\frac{1}{\overline{q_j}}\right)=\frac{a{m}^2}{n}J.$$

The defect lemma permits $\overline{z_j}$ to be replaced by $1/z_j$ and $1/\overline{q_j}$ by $q_j$. The coefficient $m/n$ can only reduce the error, so there is a complex number ${\epsilon }_J$ with

$$|{\epsilon }_J|\le 1-|J{|}^2$$

such that

$$J\sum _{j=1}^m\left(\frac{m}{n{z}_j}+q_j\right)=\frac{a{m}^2}{n}J+{\epsilon }_J.$$

Since ${\sum }_j{q}_j=m\mu$, multiplication by $n/m$ gives the useful form

$$J\sum _{j=1}^m\frac{1}{z_j}=amJ-n\mu J+\frac{n}{m}{\epsilon }_J.$$

On the other hand, the two origin identities say

$${\int }_0^{1}F(t)dt=\frac{(-1{)}^{m}J}{n}$$

and

$$F(1)=\frac{(-1{)}^{m}J}{n}\left(1+a\sum _{j=1}^m\frac{1}{z_j}\right).$$

Consequently

$$\begin{array}{r}F(1)+ma\mu {\int }_0^{1}F(t)dt=\frac{(-1{)}^m}{n}\left[(1+a^{2}m-a\mu )J+\frac{an}{m}{\epsilon }_J\right].\end{array}$$

Substitution into the integrated differential inequality therefore gives

$$\begin{array}{rl}1\le & |J|\left|\frac{a^{2}m}{n}+\frac{1}{n}-\frac{a\mu }{n}\right|+\frac{a}{m}(1-|J{|}^2)\\ & +a^{2}m{\int }_0^{1}t\beta (t{)}^{(m-1)/2}dt.\end{array}$$

For $n\ge 5$,

$$\left|\frac{a^{2}m}{n}+\frac{1}{n}-\frac{a\mu }{n}\right|\ge \frac{a^{2}m+1-a}{n}\ge \frac{2a}{m}.$$

The second inequality is equivalent to positivity of

$$m^2{a}^2-(3m+2)a+m.$$

Its discriminant $(3m+2{)}^2-4m^3$ is negative for $m\ge 4$, so the quadratic is positive for every real $a$. If

$$A_J:=\left|\frac{a^{2}m}{n}+\frac{1}{n}-\frac{a\mu }{n}\right|,$$

then $A_J\ge 2a/m$ and

$$\frac{d}{dr}\left(r{A}_J+\frac{a}{m}(1-r^2)\right)=A_J-\frac{2ar}{m}\ge 0(0\le r\le 1).$$

Thus replacing $|J|$ by $1$ can only increase this part of the right side, and the preceding inequality implies

$$1\le \left|\frac{a^{2}m+1-a\mu }{n}\right|+a^{2}m{\int }_0^{1}t\beta (t{)}^{(m-1)/2}dt.$$

Finally, for $s>0$,

$$|s+it|\le s+\frac{t^2}{2s}.$$

For

$$W:=\frac{a^{2}m+1-a\mu }{n},$$

the real part

$$s=\frac{a^{2}m+1-ax}{n}$$

is at least $a^{2}m/n>0$, while the imaginary part is $-ay/n$. Hence

$$\begin{array}{rl}|W|& \le \frac{a^{2}m+1-ax}{n}+\frac{y^2}{2nm}\\ & \le 1-\frac{2\alpha +ax}{n}+\frac{1-x^2}{2nm}.\end{array}$$

Insert this bound above, cancel $1$, and multiply by $n$. Since $(m-1)/2=(n-2)/2$, one obtains the raw origin inequality

$$2\alpha +ax\le \frac{1-x^2}{2m}+a^{2}nm{\int }_0^{1}t\beta (t{)}^{(n-2)/2}dt.$$

This estimate has two useful consequences.

A Uniform Bound on the Boundary Scale

Boundary-scale bound

Every hypothetical counterexample with $n\ge 5$ satisfies

$$\alpha \le 17.$$

Assume $\alpha >17$. Since $\alpha <m/2$, this forces $m\ge 35$ and $n\ge 36$. The quadratic $\beta$ has derivative $2a(at-x)$ and its minimum at $t=x/a$. On $0\le t\le x/a$, convexity places its graph below the chord joining $(0,1)$ to the minimum, so

$$\beta (t)\le 1-axt\le e^{-axt}.$$

If $x/a\le 1$, then $\beta$ is increasing on $[x/a,1]$ and hence $\beta (t)\le b$ there; if $x/a>1$, this second interval is empty. Therefore

$${\int }_0^{1}t\beta (t{)}^{(n-2)/2}dt\le \frac{4}{(n-2{)}^2{a}^2{x}^2}+\frac{1}{2}{b}^{(n-2)/2}.$$

The two polar bounds imply

$$\frac{1}{x^2}\le \frac{\alpha }{\log \alpha },b\le {\alpha }^{-1/\alpha }.$$

Indeed, $x^2\ge 1-b$ and the logarithmic polar bound gives

$$x^2\ge 1-b>\frac{\log \alpha }{\alpha }.$$

Also, $0<b<1$ and $\log b\le b-1$ imply

$$b\le e^{-(1-b)}<e^{-\log (\alpha )/\alpha }={\alpha }^{-1/\alpha }.$$

Put

$$v=\frac{a^2}{1-a^2}.$$

Then

$$a^2=\frac{v}{1+v},\alpha =\frac{m}{2(1+v)},\frac{n-2}{2\alpha }=1+v-\frac{1}{2\alpha }.$$

Discarding the nonnegative term $ax$ from the left side of the raw origin inequality, inserting the integral bound, and dividing by $\alpha$ gives

$$2\le \frac{1}{2m\alpha }+\frac{4nm}{(n-2{)}^2\log \alpha }+\frac{2n}{m}v(1+v){\alpha }^{-v}{\alpha }^{1/(2\alpha )}.$$

For completeness, the function

$$g(v):=v(1+v)17^{-v}$$

has logarithmic derivative

$$\frac{g^{\prime }(v)}{g(v)}=\frac{1}{v}+\frac{1}{1+v}-\log 17.$$

It has a unique positive critical point

$$v_{\ast }=\frac{2-\log 17+\sqrt{(\log 17{)}^2+4}}{2\log 17},$$

which is its maximum. Elementary enclosures for $\log 17$ give

$$g(v)\le g(v_{\ast })<\mathrm{0.1825.}$$

The remaining factors are largest at $n=36$ and $\alpha =17$: $n/(n-1)$ and $n(n-1)/(n-2{)}^2$ decrease with $n$, while ${\alpha }^{1/(2\alpha )}$ decreases for $\alpha \ge 17$. Consequently the right side is at most

$$\frac{1}{2\cdot 35\cdot 17}+\frac{4\cdot 36\cdot 35}{34^2\log 17}+\frac{72}{35}(0.1825)17^{1/34}<1.948,$$

contradicting the left side $2$.

A Scalar Form of the Origin Inequality

Put $r=(n-2)/2$. The decomposition

$$\beta (t)=(1-axt{)}^2+a^2{t}^2(1-x^2)$$

has two nonnegative terms because $1-axt>0$. Since $r\ge 3/2$, the mean value theorem gives

$$\begin{array}{rl}\beta (t{)}^{(n-2)/2}\le & (1-axt{)}^{n-2}\\ & +\frac{n-2}{2}{a}^2{t}^2(1-x^2)\beta (t{)}^{(n-4)/2}.\end{array}$$

The first term is integrated by extending the interval from $[0,1]$ to $[0,1/(ax)]$. With $u=axt$,

$${\int }_0^{1/(ax)}t(1-axt{)}^{n-2}dt=\frac{1}{a^2{x}^2}{\int }_0^{1}u(1-u{)}^{n-2}du=\frac{1}{a^2{x}^{2}n(n-1)}.$$

Thus the raw origin inequality implies

$$\begin{array}{rl}2\alpha +ax\le & \frac{1-x^2}{2m}+\frac{1}{x^2}\\ & +\frac{a^{4}nm(n-2)(1-x^2)}{2}{\int }_0^1{t}^3\beta (t{)}^{(n-4)/2}dt.\end{array}$$

Use also

$$1-x^2\le b,\frac{1}{x^2}\le 1+\frac{b}{1-b},ax=1-\frac{\alpha }{m}-\frac{b}{2}.$$

The three displayed relations follow respectively from $b\ge 1-x^2$, from $x^2\ge 1-b>0$, and from the definitions of $\alpha$ and $b$. Substitute them into the preceding inequality, move $ax$ to the right, and divide by $2\alpha$ to obtain

$$\begin{array}{rl}1\le & \frac{b}{2\alpha (1-b)}+\frac{b}{4\alpha }+\frac{1}{2m}+\frac{b}{4\alpha m}\\ & +\frac{a^{4}nm(n-2)b}{4\alpha }{\int }_0^1{t}^3\beta (t{)}^{(n-4)/2}dt.\end{array}$$

🏷️ Incompatibility of the Two Channels

For fixed $\alpha$, use

$$ax=1-\frac{\alpha }{m}-\frac{b}{2}$$

to regard

$$\beta (t)=1-2\left(1-\frac{\alpha }{m}-\frac{b}{2}\right)t+a^2{t}^2$$

as a function of $b$. It is nondecreasing in $b$ for every $t\in [0,1]$, and every other occurrence of $b$ on the right side of the scalar origin inequality is also increasing on $0<b<1$. The polar constraint therefore permits the replacement

$$b<\frac{\alpha }{3+\alpha }.$$

Set

$$A:=a^2=1-\frac{2\alpha }{m},$$ $$c:=1-\frac{\alpha }{m}-\frac{\alpha }{2(3+\alpha )},$$

and

$$Q_{n,\alpha }(t):=1-2ct+A{t}^2.$$

The inequality $\alpha <m/2$ gives $c>0$, because its first two terms have difference greater than $1/2$ while $\alpha /(2(3+\alpha ))<1/2$. Moreover $b<\alpha /(3+\alpha )$ gives $c<ax$, and $x\le 1$ gives $ax\le a$. Hence

$$0<c<ax\le a,c^2\le A.$$

Every counterexample of degree $n\ge 5$ would therefore satisfy

$$0<\alpha \le 17,c^2\le A,$$

and

$$1\le R_n(\alpha ),$$

where

$$\begin{array}{rl}{R}_n(\alpha ):=& \frac{1}{6}+\frac{1}{4(3+\alpha )}+\frac{1}{2m}+\frac{1}{4m(3+\alpha )}\\ & +\frac{A^{2}nm(n-2)}{4(3+\alpha )}{\int }_0^1{t}^3{Q}_{n,\alpha }(t{)}^{(n-4)/2}dt.\end{array}$$

The remaining assertion is purely real and one-dimensional:

$$R_n(\alpha )<1$$

throughout the feasible set. All polynomial zero geometry has disappeared.

An Elementary Large-Degree Estimate

The blog gives a short closure for $n\ge 201$. On $0\le t\le c/A$, one has $A{t}^2\le ct$ and therefore

$$Q_{n,\alpha }(t)\le 1-ct\le e^{-ct}.$$

Here $1-ct\ge 0$ follows from $c^2\le A$. After its vertex $c/A$, the quadratic increases, so its maximum before $t=1$ is

$$Q_{n,\alpha }(1)=\frac{\alpha }{3+\alpha }.$$

If the vertex lies beyond $1$, the second interval is simply empty. Thus

$$Q_{n,\alpha }(t)\le \{\begin{array}{ll}{e}^{-ct},& 0\le t\le c/A,\\ \alpha /(3+\alpha ),& c/A<t\le 1.\end{array}$$

Consequently,

$$\begin{array}{rl}{\int }_0^1{t}^3{Q}_{n,\alpha }(t{)}^{(n-4)/2}dt\le & \frac{96}{(n-4{)}^4{c}^4}\\ & +\frac{1}{4}{\left(\frac{\alpha }{3+\alpha }\right)}^{(n-4)/2}.\end{array}$$

For $n\ge 201$ and $0\le \alpha \le 17$,

$$(3+\alpha )c^4\ge 1.$$

Indeed, $c$ increases with $n$, so it suffices to take $n=201$. The derivative of $(3+\alpha )c^4$ is then

$$-c^3\frac{5(3+\alpha {)}^2-103(3+\alpha )+900}{200(3+\alpha )},$$

which is negative because the quadratic numerator has discriminant $-7391$. The minimum occurs at $\alpha =17$, where

$$(3+17)c^4=20{\left(\frac{49}{100}\right)}^4>1.$$

It follows that

$$\begin{array}{rl}{R}_n(\alpha )\le & \frac{1}{6}+\frac{1}{12}+\frac{7}{12(n-1)}+\frac{24n(n-1)(n-2)}{(n-4{)}^4}\\ & +\frac{n(n-1)(n-2)}{48}{\left(\frac{17}{20}\right)}^{(n-4)/2}.\end{array}$$

Every $n$-dependent term decreases for $n\ge 201$. For the polynomial tail this follows from

$$\frac{n+1}{n-2}{\left(\frac{n-4}{n-3}\right)}^4<1,$$

and for the exponential tail from

$$\frac{n+1}{n-2}\sqrt{\frac{17}{20}}<\frac{202}{199}\frac{12}{13}<1.$$

At $n=201$, the rational estimate $\sqrt{17/20}<12/13$ gives

$$\begin{array}{rl}{R}_{201}(\alpha )<& \frac{1}{6}+\frac{1}{12}+\frac{7}{2400}+\frac{24\cdot 201\cdot 200\cdot 199}{197^4}\\ & +\frac{201\cdot 200\cdot 199}{48}{\left(\frac{17}{20}\right)}^{98}\frac{12}{13}<\frac{2}{5}.\end{array}$$

This contradicts $R_n(\alpha )\ge 1$ for all $n\ge 201$.

The Finite Exact Check

The bounded range is not settled in the blog by a single displayed hand inequality. The computation is nevertheless exact and algebraic rather than floating-point evidence. The key observations are

$$Q_{n,\alpha }(t)=(1-ct{)}^2+(A-c^2)t^2\ge 0$$

and, by convexity with $Q(0)=1$ and $Q(1)=\alpha /(3+\alpha )$,

$$0\le Q_{n,\alpha }(t)\le 1.$$

For an integer $k\ge 0$, the moment is the exact finite sum

$$\begin{array}{rl}{I}_k(n,\alpha )& :={\int }_0^1{t}^3{Q}_{n,\alpha }(t{)}^{k}dt\\ & =\sum _{\begin{array}{c}i,j\ge 0\\ i+j\le k\end{array}}\frac{k!}{i!j!(k-i-j)!}\frac{(-2c{)}^i{A}^j}{4+i+2j}.\end{array}$$

For even $n$, this is the required integral with $k=(n-4)/2$. For odd $n\ge 5$, put $k=(n-5)/2$. For every rational $w>0$, the square $(\sqrt{Q}-w{)}^2\ge 0$ gives the tangent bound

$$\sqrt{Q}\le \frac{Q}{2w}+\frac{w}{2}$$

and hence

$$Q^{k+1/2}\le \frac{Q^{k+1}}{2w}+\frac{w}{2}{Q}^k.$$

The current certificate construction takes $w=1/3$, so the odd moment is bounded by

$$\frac{3}{2}{I}_{k+1}(n,\alpha )+\frac{1}{6}{I}_k(n,\alpha ).$$

This is sharper than the universal choice $w=1$, particularly in degree five. Thus each fixed degree reduces to a rational inequality in $\alpha$. After moving everything to the left and clearing a known positive denominator, write the desired inequality as $P_n(\alpha )>0$.

On a rational interval $[\ell ,u]$, write

$$P_n(\ell +(u-\ell )s)=\sum _{j=0}^d{\gamma }_j\left(\genfrac{}{}{0ex}{}{d}{j}\right)s^j(1-s{)}^{d-j}.$$

The Bernstein basis is nonnegative and sums to one on $0\le s\le 1$. Hence

$$\underset{j}{\min }{\gamma }_j>0⟹P_n(\alpha )>0(\ell \le \alpha \le u).$$

All endpoints and coefficients are rational, so this is an exact certificate, not a sampling argument. The relevant interval can be taken as

$$0\le \alpha \le \min \left\{17,\frac{n-1}{2}\right\};$$

the feasibility hypothesis $c^2\le A$ guarantees $Q\ge 0$. The generator may discover the coefficients, but a checker must re-establish both the polynomial identity and the strict positivity of every coefficient using exact integer or rational arithmetic.

Tao’s original blog split used a finite check through degree $200$ and the elementary estimate above afterward. The current companion repository improves the division of labor: rational Bernstein certificates handle degrees $5$ through $100$, while a sharper uniform analytic tail begins at degree $101$ (Tao, 2026b). Here “analytic tail” means uniform in $n$, not entirely computation-free: its final one-variable endpoint estimate is also closed by one exact rational Bernstein certificate. The public documentation records that the odd-degree path uses the tangent parameter $w=1/3$ and that no floating-point number is a proof input. The original Mazur proof uses a different scalar reduction and different exact certificate objects (Mazur, 2026).

Regime	Closure of the argument
$2\le n\le 5$	the weighted AM—GM hand proof below
$5\le n\le 100$	exact rational Bernstein certificates; degree five is an intentional overlap
$n\ge 101$	a uniform analytic reduction and one exact rational endpoint certificate
$|a|=1$	the hand proof of the Rubinstein equality case

Meaning of “elementary”

The structural proof uses polynomial factorization, a disk Möbius map, AM—GM, Cauchy—Schwarz, Maclaurin’s inequality, elementary calculus, and power-series positivity. It does not use contour integration, potential theory, or the asymptotic machinery of earlier high-degree proofs. It is therefore elementary in method. The currently documented proof is still computer-assisted because a finite family of large rational positivity certificates is part of the closure.

🏷️ Degrees Two Through Five

The low degrees do not require the origin channel. From the polar lower bound and $|q_j|\le 1$,

$$1\le {\int }_0^1(a+(1-a^2)t{)}^{m}dt.$$

Put $X(t)=a+(1-a^2)t$. A direct calculation gives

$$1-{\int }_0^{1}X(t{)}^{4}dt=\frac{(1-a{)}^3(1+a)}{5}(a^4-3a^3+3a+4)>0.$$

One can obtain the factorization by first writing

$${\int }_0^{1}X(t{)}^{4}dt=\frac{(1+a-a^2{)}^5-a^5}{5(1-a^2)}$$

and then clearing the positive factor $1-a^2$.

The last factor is positive because

$$a^4-3a^3+3a+4=2+3a+(1-a)(2+2a+a^2(2-a)).$$

For $1\le m\le 4$, put $\theta =m/4$. Weighted AM—GM applied to the two nonnegative numbers $1$ and $X(t{)}^4$ gives pointwise

$$X(t{)}^m=1^{1-m/4}(X(t{)}^4{)}^{m/4}\le 1-\frac{m}{4}+\frac{m}{4}X(t{)}^4.$$

After integration,

$${\int }_0^{1}X(t{)}^{m}dt<1,$$

contradicting the polar lower bound. This proves the interior statement for $2\le n\le 5$.

No estimate $X(t)\le 1$ is being used; indeed $X(1)=1+a-a^2$ can exceed one. The strictness comes only from the exact integral of $X^4$ and from $m>0$.

🏷️ Boundary Zeroes and the Equality Case

It remains to treat $|a|=1$. Rotate so that $a=1$, and assume no critical point lies at distance strictly less than one from $1$. Then $p^{\prime }(1)\ne 0$, so the distinguished zero is simple and none of the remaining zeroes $z_j$ equals $1$. Again write the critical points as

$$w_j=1-\frac{1}{q_j},0<|q_j|\le 1.$$

The polar identity degenerates because the reflected point $1/a$ coincides with $a$. Its replacement comes from $p^{\prime \prime }(1)/p^{\prime }(1)$. If

$$p(z)=c(z-1)g(z),g(z)=\prod _{j=1}^m(z-z_j),$$

then

$$\frac{p^{\prime \prime }(1)}{p^{\prime }(1)}=2\frac{g^{\prime }(1)}{g(1)}=2\sum _{j=1}^m\frac{1}{1-z_j}.$$

On the critical-point side,

$$p^{\prime }(z)=cn\prod _{j=1}^m(z-w_j)$$

gives

$$\frac{p^{\prime \prime }(1)}{p^{\prime }(1)}=\sum _{j=1}^m\frac{1}{1-w_j}=\sum _{j=1}^m{q}_j.$$

Equating these expressions gives the Meir—Sharma identity

$$\sum _{j=1}^m{q}_j=2\sum _{j=1}^m\frac{1}{1-z_j}.$$

For $|z_j|\le 1$ and $z_j\ne 1$,

$$\mathrm{Re}\frac{1}{1-z_j}-\frac{1}{2}=\frac{1-|z_j{|}^2}{2|1-z_j{|}^2}\ge 0.$$

The real part of the right side is therefore at least $m$, while

$$\mathrm{Re}\sum _j{q}_j\le \sum _j|q_j|\le m.$$

Because the two sides are equal, both real-part bounds must be equalities. Each number $1-\mathrm{Re}{q}_j$ is nonnegative and their sum is zero, so $\mathrm{Re}{q}_j=1$ for every $j$. Together with $|q_j|\le 1$, this forces $q_j=1$. Every critical point is therefore $0$, and

$$p^{\prime }(z)=cn{z}^{n-1}.$$

Since $p(1)=0$,

$$p(z)=c(z^n-1).$$

Undoing the rotation gives $p(z)=c(z^n-a^n)$ for an arbitrary boundary zero $a$. This is Rubinstein’s equality classification (Rubinstein, 1968); the short reciprocal proof is also implicit in the translated identity used by Tang and Zhang (Tang & Zhang, 2025). Together with the strict interior result, it proves Phelps—Rodriguez and hence Sendov.

🏷️ Structural Interpretation

The proof can be compressed into the following chain:

$$\begin{array}{c}\text{hypothetical counterexample}\\ \Downarrow \\ z_j,q_j\in \overline{\mathbb{D}}\text{ satisfying four identities}\\ \Downarrow \\ \text{polar upper bound on }b=1-2ax+a^2\\ \Downarrow \Downarrow \\ \alpha \le 17\text{origin lower bound on }b\\ \Downarrow \\ 1\le R_n(\alpha )<1.\end{array}$$

The unexpected feature is the amount of slack. The proof discards almost all information determining critical points from zeroes, enlarging the apparent degrees of freedom, yet the surviving constraints are already incompatible. Maclaurin’s inequality is the compression mechanism: a codimension-one symmetric sum of $m$ factors is controlled by the single quadratic statistic $\beta (t)$. The defect lemma then quantifies how far inversion inside the disk deviates from conjugation on its boundary.

This explains both why the proof is elementary and why it was difficult to find. The hard step was not deploying a powerful theorem; it was selecting the reciprocal critical coordinates, the polar evaluation $1/a$, the two origin evaluations, and the correct two real summary statistics. Once those choices are made, the remaining proof has considerable numerical room except in the narrow boundary layer $a=1-O(1/n)$.

See Also

• on random polynomial zeros --- contrasts the deterministic geometry linking zeroes and critical points here with probabilistic zero statistics and local first-zero certificates.
• on two proofs of Crouzeix’s conjecture --- another 2026 resolution in complex analysis where the decisive advance is preserving a small amount of structure that earlier norm reductions discarded.

📚 References

🐻  Brown, J.E. & Xiang, G. 1999. Proof of the Sendov Conjecture for Polynomials of Degree at Most Eight. Journal of Mathematical Analysis and Applications 232(2), 272–292.

🐻  Hayman, W.K. 1967. Research Problems in Function Theory, London: Athlone Press,p.

🐻  Mazur, L. 2026. A Computer-Assisted Proof of Sendov’s Conjecture.

🐻  Phelps, D. & Rodriguez, R.S. 1972. Some Properties of Extremal Polynomials for the Ilieff Conjecture. Kodai Mathematical Seminar Reports 24, 172–175.

🐻  Rubinstein, Z. 1968. On a Problem of Ilyeff. Pacific Journal of Mathematics 26, 159–161.

🐻  Tang, Q. & Zhang, T. 2025. Sharp Schoenberg Type Inequalities and the de Bruin–Sharma Problem.

🐻  Tao, T. 2022. Sendov’s Conjecture for Sufficiently-High-Degree Polynomials. Acta Mathematica 229(2), 347–392.

🐻  Tao, T. 2026a. A Digestion of the Proof of Sendov’s Conjecture.

🐻  Tao, T. 2026b. Sendov’s Conjecture in Lean.