Algebraic independence of pi and e

Algebraic independence of pi and e

Definitions and mathematical statement

Define $e=\sum_{n=0}^{\infty}1/n!$ and let $\pi$ be the positive half-period of the usual sine function. Algebraic independence of $e$ and $\pi$ over ${\mathbb{Q}}$ means \[ \forall P\in{\mathbb{Q}}[X,Y]\setminus\{0\},\qquad P(e,\pi)\ne0. \] Equivalently, the evaluation homomorphism ${\mathbb{Q}}[X,Y]\to{\mathbb{R}}$ is injective, or the transcendence degree of ${\mathbb{Q}}(e,\pi)$ over ${\mathbb{Q}}$ is two. This is stronger than showing either number individually transcendental or showing just $e+\pi$ irrational. A counterexample would require a nonzero polynomial relation with rational coefficients, not a numerical near-equality.

Formulation and concept references: [S1].

Short English statement

Are the constants e and pi free of every nontrivial polynomial relation with rational coefficients?

Research attempt

Current frontier.

Individual transcendence, algebraic independence of some other exponential constants, and functional transcendence must be separated. Pila's notes state the $e,\pi$ problem and explain Schanuel's conjecture and functional analogues; the cited Nesterenko result involves $\pi$ and $e^\pi$, not $e$ and $\pi$ [S1], Sections 2--4. A 2026 manuscript studies a symmetry between recurrences associated with $e$ and $\pi$ and their meromorphic connection functions [E1], Sections 1 and 9. Such a relationship between functions or limiting recurrences is not a polynomial relation between the two numbers. No unconditional solution of the printed target was verified in the current source search. The work below combines an algebraic reformulation, an auxiliary-function attempt, and an exact finite nonvanishing certificate.

Attack route.

A conditional transcendence-degree argument is very short but imports Schanuel's conjecture. A functional approach would need an arithmetic specialization theorem that survives evaluation at the particular arguments $1$ and $i\pi$. An auxiliary-function approach would need integer linear forms in mixed monomials with simultaneously controlled size, denominators, and nonvanishing. A search for a counterexample must use exact certification rather than a numerical near-relation. Each route is pursued far enough to expose its separate missing condition.

Attempt: symmetric coordinates lose no transcendence degree.

Put $s=e+\pi$ and $t=e\pi$. Since $e$ and $\pi$ are roots of $Z^2-sZ+t$, the field ${\mathbb{Q}}(e,\pi)$ is algebraic over ${\mathbb{Q}}(s,t)$, and \begin{equation}\label{wsref-0-33-33-34-39-3a-73-79-6d-6d-65-74-72-69-63} \operatorname{trdeg}_{{\mathbb{Q}}}{\mathbb{Q}}(e,\pi) =\operatorname{trdeg}_{{\mathbb{Q}}}{\mathbb{Q}}(s,t). \end{equation} More concretely, if a nonzero $P(X,Y)\in{\mathbb{Q}}[X,Y]$ vanished at $(e,\pi)$, then $P(X,Y)P(Y,X)$ would be a nonzero symmetric polynomial. The fundamental theorem of symmetric polynomials writes it uniquely as $Q(X+Y,XY)$ for a nonzero $Q\in{\mathbb{Q}}[S,T]$. Thus every proposed relation produces one in the symmetric coordinates, though possibly of higher degree and height.

This does not show that both $s$ and $t$ are transcendental. It does show that they cannot both be algebraic, since then $e$ and $\pi$ would both be algebraic, contradicting their known individual transcendence [S1], Section 2. That elementary consequence is strictly weaker than (3349:symmetric) having value two, and even an irrationality proof for $s$ alone would not establish the target.

Attempt: the precise Schanuel dependency.

The numbers $1$ and $i\pi$ are linearly independent over ${\mathbb{Q}}$, since the real and imaginary parts of $a+bi\pi=0$ give $a=b=0$. Schanuel's conjecture for just this pair would imply \begin{equation} \operatorname{trdeg}_{{\mathbb{Q}}}{\mathbb{Q}}(1,i\pi,e,e^{i\pi})\geq2. \end{equation} Euler's identity gives $e^{i\pi}=-1$. Adjoining $i$ is algebraic and does not change transcendence degree, so the conclusion would be $\operatorname{trdeg}_{{\mathbb{Q}}}{\mathbb{Q}}(e,\pi)=2$. This is the familiar conditional route in the framework of [S1], Section 3, not a proof of the needed numerical instance of Schanuel. The full target also implies transcendence of both $e+\pi$ and $e\pi$, by substituting $X+Y$ or $XY$ into a proposed algebraic equation. These consequences are useful warnings against treating a new "specialization lemma" as innocuous.

Attempt: why functional independence does not specialize.

For an irrational real $\alpha$, $\exp z$ and $\exp(\alpha z)$ are algebraically independent as analytic functions over ${\mathbb{C}}$. A polynomial relation would be a finite linear combination of $\exp((m+n\alpha)z)$ with distinct exponents. Differentiating at zero produces a Vandermonde system, so every coefficient vanishes. Yet at $z=0$ both functions equal one. This proves, without any transcendence conjecture, that even exact functional algebraic independence can have an algebraically dependent special value.

The functional results discussed in [S1], Sections 4--6, control positive-dimensional analytic or algebraic configurations. Applying one directly to the isolated pair of arguments $(1,i\pi)$ would require an additional arithmetic statement; the functional result alone does not supply it. The same problem affects a proposed use of the recurrence symmetries in [E1]: relations among connection functions have to be accompanied by a theorem about these particular values, not just their functional equations.

Attempt: a mixed auxiliary-function calculation.

Fix an integer $d\geq1$ and seek \begin{equation} F(z)=\sum_{0\leq i,j\leq d}c_{ij}z^i e^{jz}, \qquad c_{ij}\in{\mathbb{Z}}, \end{equation} with a large zero at $z=0$. There are $M=(d+1)^2$ coefficients. Vanishing of the first $M-1$ Taylor coefficients gives $M-1$ rational homogeneous linear equations, so a nonzero rational solution exists; clearing denominators gives an integral one. The functions $z^ie^{jz}$ are linearly independent: on the positive real axis, divide a putative identity by the largest exponential and let $z\to+\infty$; its leading polynomial would tend to zero, so must vanish. Induction removes all exponentials. Thus this construction gives a genuinely nonzero $F$.

Its two evaluations have sharply different arithmetic forms: \begin{equation}\label{wsref-0-33-33-34-39-3a-65-6e-64-70-6f-69-6e-74-73} F(1)=\sum_{j=0}^d\left(\sum_i c_{ij}\right)e^j, \qquad F(i\pi)=\sum_{i=0}^d\left(\sum_j(-1)^jc_{ij}\right)(i\pi)^i. \end{equation} The second expression uses exactly $e^{ji\pi}=(-1)^j$. In this maximal-order construction one can in fact prove that both collected polynomials in (3349:endpoints) are nonzero. Here is a useful check that goes beyond merely observing $F\not\equiv0$. An exponential polynomial of degrees $i\le u$, $j\le v$ is annihilated by the constant-coefficient differential operator \begin{equation} \prod_{j=0}^v\left(\frac{\mathrm d}{\mathrm dz}-j\right)^{u+1}, \end{equation} of order $(u+1)(v+1)$. If its first that many derivatives vanish at zero, the differential equation recursively forces every derivative to vanish, so the function is identically zero.

Write $P_0(Z,W)=\sum c_{ij}Z^iW^j$. If the first polynomial in (3349:endpoints) were identically zero, then $P_0(1,W)=0$ and $P_0$ would be divisible by $Z-1$. Dividing $F(z)$ by $z-1$ would give an exponential polynomial with degree bounds $(d-1,d)$ and the same zero of order at least $M-1$ at zero. Its annihilating operator has order $d(d+1)\le M-1$, an impossibility. If the second polynomial were identically zero, then $P_0(Z,-1)=0$ and $P_0$ would be divisible by $W+1$. Dividing $F(z)$ by $e^z+1$, which is nonzero at zero, similarly gives degree bounds $(d,d-1)$ and the same contradiction. Thus both endpoint polynomials are nonzero; individual transcendence of $e$ and $\pi$ now proves $F(1)F(i\pi)\ne0$ [S1], Section 2. This is nonvanishing of these separated forms, not algebraic independence of all mixed monomials. The script independently checks the exact rational interpolation matrix, zero order, and nonzero collected endpoint polynomials for $1\le d\le4$; the argument above covers every $d$.

There is also an explicit coefficient bound. Use derivative equations rather than Taylor coefficients: their matrix entries are $(r!/(r-i)!)j^{r-i}$ for $r\ge i$, and zero otherwise, for $0\le r\le M-2$, with $0^0=1$. Its rank is $M-1$ by the same annihilating-operator argument. Taking maximal minors gives a nonzero integer kernel vector satisfying, for $d\ge2$, \begin{equation} H:=\max|c_{ij}| \le\left(\sqrt{M-1}\,M^d d^{M-2}\right)^{M-1}. \end{equation} This is Hadamard's inequality applied to the minor matrices. If $m=M-1$ and $\rho>|z|$, the maximum principle for $F(z)/z^m$ gives \begin{equation} |F(z)|\le M H\max(1,\rho^d)e^{d\rho} \left(\frac{|z|}{\rho}\right)^m. \end{equation} Choosing $\rho=m/d$ makes the negative logarithmic contribution at a fixed endpoint of scale $-d^2\log d$, but the available upper bound on $\log H$ has scale $d^4\log d$. Substituting this coefficient bound therefore does not even force these endpoint values to be small. This failure is about the bounds proved here; it does not show that sharper auxiliary functions or sharper coefficient estimates are impossible.

To derive a contradiction from an arbitrary $P(e,\pi)=0$, one still needs an arithmetic lower bound for suitable mixed forms, with coefficient and denominator growth dominated by analytic smallness. The separated nonvanishing lemma supplies no such mixed estimate. If maximal-order interpolation is discarded, the naive nonvanishing principle fails even more directly: $(e^z-1)^m(e^z+1)$ is nonzero as a function, vanishes to order $m$ at zero, and vanishes at $i\pi$. Thus neither a Taylor remainder nor high-order vanishing by itself is an arithmetic independence theorem.

Exact computation: a bounded polynomial family.

The accompanying standard-library Python script uses rational interval arithmetic, not floating-point root recognition. It proves the finite statement \begin{equation}\label{wsref-0-33-33-34-39-3a-66-69-6e-69-74-65} \left|P(e,\pi)\right|>2^{-10} \quad\begin{gathered} \text{for every nonzero }P\in{\mathbb{Z}}[X,Y]\text{ of total degree at most }3,\\ \text{with every coefficient in }\{-1,0,1\}. \end{gathered} \end{equation} There are $3^{10}-1=59\,048$ such polynomials. Set $E_N=\sum_{j=0}^N1/j!$. A geometric-series comparison gives \begin{equation} E_N<e<E_N+\frac1{N\,N!}\qquad(N\geq1). \end{equation} For $0<x<1$, consecutive even and odd truncations of $\arctan x=\sum_{j\geq0}(-1)^j x^{2j+1}/(2j+1)$ enclose its value. Machin's identity \begin{equation} \pi=16\arctan(1/5)-4\arctan(1/239) \end{equation} follows by the tangent addition formula and the fact that $4\arctan(1/5)-\arctan(1/239)$ lies in $(0,\pi/2)$ with tangent one. The factorial sum through $1/80!$ and eighty arctangent terms suffice for the following dyadic outer enclosures. If $D=2^{128}$, then \begin{align*} L_e&=924983374546220337150911035843336795079, & L_e&<De<L_e+1,\\ L_\pi&=1069028584064966747859680373161870783300, & L_\pi&<D\pi<L_\pi+1. \end{align*} For readability the script, rather than a table of decimal values, records the exact integers and the assertion that the analytic intervals lie inside these outer bounds.

All monomials are positive at the endpoints. Bound $e^i\pi^j$ by the products of the lower and upper endpoint powers, and reverse endpoints when a coefficient is negative. Put every result over the common denominator $D^3$, including terms of smaller degree. Exhaustive integer addition verifies that each polynomial interval lies entirely outside $[-2^{-10},2^{-10}]$. The polynomial attaining the smallest certified distance in this enumeration is \begin{equation} P_*(X,Y)=X^3-Y^3+X^2-Y^2+XY+X+Y-1 \end{equation} or its negative. Its closeness to zero is not a relation; the interval proof explicitly separates it from zero. This is a finite, reproducible nonvanishing proof for (3349:finite), not a probabilistic search claim. No assertion is made for other degrees or coefficient heights.

Stress test and outcome.

Outcome: Exploratory approach with unresolved gap.

The symmetric reformulation is exact, the Schanuel implication is explicitly conditional, and the auxiliary-function calculation proves separated endpoint nonvanishing but stops at an uncontrolled mixed arithmetic estimate. The computation proves a bounded-family certificate with rational error control; it does not promote a finite search or a very small decimal to an algebraic independence theorem. The universal quantifier over all nonzero rational polynomials, equivalently all integral polynomials after clearing denominators, remains unhandled. No counterexample polynomial or unconditional complete proof candidate has been obtained.

Sources

• S1 Jonathan Pila, Functional transcendence via o-minimality, edited LMS-EPSRC 2013 lecture notes. https://people.maths.ox.ac.uk/pila/LMSNotes.pdf. Catalogue formulation source.
• E1 Benoit Cloitre, "The holonomic triangle: from a symmetry between $e$ and $\pi$ to additive Gamma functions," arXiv:2601.04242v1 (January 2026). Sections 1 and 9. https://arxiv.org/pdf/2601.04242. Recent recurrence/functional connection, not a polynomial relation or an independence proof. Only the stated scope is used.