"Almost true": non-trivial claims that have exactly one counterexample

Question

Question:

What are some non-trivial claims that have exactly one counterexample?

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Example 1:

"There is no Pythagorean triple whose numbers are consecutive terms in a row of Pascal's triangle." There is exactly one counterexample: $\left\{\binom{62}{26},\binom{62}{27},\binom{62}{28}\right\}$. proof

Example 2:

"If $n^2>1$ cannonballs are laid on the ground in a filled square formation, then they cannot all be used to make a square pyramid of cannonballs." There is exactly one counterexample: $n=70$. (from this answer)

Example 3:

"No number is equal to the sum of the subfactorials of its digits." There is exactly one counterexample: $148349$. (from this answer)

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Remarks:

• It must be proven that there exists exactly one counterexample.
• Let's exclude trivially false claims, i.e. claims that are obviously false, even if one has never heard the claim before, such as "There is no even prime number" (an obvious counterexample is $2$), or "For integers $n>1$, the equation $x^n+y^n=z^n$ has no integer solutions" (an obvious counterexample is $n=2$; it is not obvious that this is the only counterexample, but the claim is still obviously false). (Note: I added the second example after a lot of answers were already given, to clarify this restriction.)
• Let's exclude ad hoc claims, i.e. claims that are purposely designed to have exactly one counterexample, such as "There is no integer $n$ such that $n=\frac{600}{\pi^2}\sum\limits_{k=1}^\infty\frac{1}{k^2}$" (the counterexample is n=100). That is, the claims should be reasonably natural claims.

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This question was inspired by the question "Conjectures that have been disproved with extremely large counterexamples?".

Answer 1

$\mathbb{R}^n$ has a unique differential structure except for $n=4$ where there are an uncountably infinite number (Wikipedia). I like this because the exception is very wrong.

Answer 2

I am fond of the fact that the automorphisms of $S_n$ are always inner automorphisms, with the sole exception of $n=6$.

Answer 3

Cannonball problem

No distinct positive integers $(M, N)$ satisfy $$ \frac16N(N+1)(2N+1) = M^2,$$ except for $(M,N)=(70,24)$. One might argue that this is the reason why Leech lattice exists and has such outstanding properties.

Answer 4

For $n \geq 0$, the commutator subgroup of the general linear group $\mathrm{GL}_n(K)$ is always the special linear group $\mathrm{SL}_n(K)$, with the exception of $n=2$ and $K=\mathbb{F}_2$. Proof.

Answer 5

There are no positive integers $a \neq b$ such that $a^b = b^a$, with the exception of $2^4=4^2$.

Answer 6

Catalan's conjecture

The only solution in the natural numbers for $x^a-y^b=1$ is $x=b=3,\; y=a=2$.

which can be stated in this form to possibly fit as an answer:

There are no consecutive powers of natural numbers, except 8 and 9.

Answer 7

Almost true: no number is equal to the sum of the subfactorials of its digits.

The only counterexample is $148349=!1+!4+!8+!3+!4+!9$.

The subfactorial $!n$ is the number of derangements of a set of $n$ elements, that is, the number of permutations $\rho$ of the set with no $x$ such that $\rho(x)=x$. Values are tabulated, with scads of information, at https://oeis.org/A000166. The result is attributed to Madachy, J. S., Madachy's Mathematical Recreations. New York: Dover, p. 167, 1979.

Answer 8

Zsigmondy's theorem states that if $a > b > 0$ are coprime integers, then for any integer $n > 2$, there is a prime number $p$ dividing $a^n - b^n$ and not dividing $a^k - b^k$ for all positive integers $k < n$, with the sole exception of $a=2, b=1, n=6$.

Answer 9

Off the top of my head, one distinct problem (distinct from the answers already given) comes to mind. Read it on this site only a while back.

Let $A$ be an $n\times n$ matrix with non-negative entries. Let $\gamma(A)$ denote the smallest positive integer $r$ (if it exists) such that $A^r$ has only positive entries.

It was conjectured that for every $n$ and every $r<1+(n^2-2n+2)/2$, there is an $n\times n$ matrix $A$ with $\gamma(A)=r$.

Zhang(1987) proved that $n=11$ is the only exception to this conjecture.

Answer 10

Background: For any associative algebra $A$, we can define a Jordan algebra $A^+$, which has the same elements as $A$ but the Jordan product $x \ast y = \frac{1}{2}(xy + yx)$. A Jordan algebra is called special if it arises as a Jordan subalgebra of $A^+$ for some $A$. It is called formally real if a sum of squares vanishes only trivially, i.e. if $\sum_{i=1}^k x_i^2 = 0$ implies $x_1 = ... = x_k = 0$.

False proposition: A finite-dimensional simple formally real Jordan algebra is special.

Unique counterexample: Let $\mathbb{A} = H_3(\mathbb{O}$) be the $3\times3$ Hermitian matrices with octonion entries, and the usual formula from above for the Jordan product. Since the Octonions are non-associative, this is not special in the obvious way, despite coming from a matrix algebra. It turns out that it is not special at all.

The uniqueness of this property is why $\mathbb{A}$ is sometimes called the exceptional Jordan algebra. It's also known as the Albert algebra.

Answer 11

This is an answer I gave years ago to another question, Accidents of small $n$.

but it works equally well here,

The alternating group $A_n$ is simple for $n\neq4$.

Answer 12

A simple Lie group is a connected non-abelian Lie group $G$ which does not have nontrivial connected normal subgroups. Let us consider the special orthogonal groups, that is, the groups of the form$$SO(n,\Bbb R)=\left\{M\in GL(n,\Bbb R)\,\middle|\,M.M^T=\operatorname{Id}_n\wedge\det M=1\right\},$$with $n\in\Bbb N$. Well, $SO(1,\Bbb R)$ and $SO(2,\Bbb R)$ are abelian, and therefore they cannot be simple. What about $SO(n,\Bbb R)$ with $n\geqslant3$? It turns out that they are all simple Lie groups, except when $n=4$.

Answer 13

Theorem 15 of Rosser and Schoenfeld's renowed paper "Approximate formulas for some functions of prime numbers", Illinois J. Math. 6 (1962) 64-94 (open access here) states that, for an integer $n \geq 3$, the following inequality holds $$\frac{n}{\varphi(n)} < e^{\gamma}\log\log n + \frac{5}{2\log\log n},$$ except when $n = 2\cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 = 223092870$, ie. when $n$ is the product of the nine first prime numbers. Here $\varphi$ is the Euler totient function and $\gamma$ is the Euler constant.

NB: one requires $n\geq 3$ since $n\leq 1$ does not make sense, and $n=2$ is trivially wrong as $\log\log 2 < 0$.

Answer 14

Almost true: there are no Carmichael numbers of the form $15pq$, $p$ and $q$ being prime numbers.

The only counterexample is $62745$.

Some explanation: a Carmichael number is a positive composite integer $n$ such that $a^{n-1}\equiv1\bmod n$ for every integer $a$ prime to $n$.

I found this statement-with-unique-counterexample in Roberts, Lure of the Integers, the section on $561$ (page 242). Roberts refers to H. J. A. Duparc, On Carmichael numbers, Simon Stevin 29 (1952) 21--24. Duparc's paper is available at https://ir.cwi.nl/pub/6931/6931D.pdf

Answer 15

Inspired by this question:

A polyhedron consisting entirely of pentagons must have an even number and at least twelve of them, a result easily verified using the Euler characteristic. But among the counts allowed by these constraints fourteen alone is impossible to reach.

A similar situation turns out to be true of polydedra with quadrangular faces. The Euler characteristic allows any number of faces greater than or equal to six, but no polyhedron exists specifically with seven quadrangular faces.

It turns out that there is a relationship between these two exceptional cases. Many readers know the graphical relationship between a cube and a regular dodecahedron. Less known is that this relationship also holds for other polyhedra: a polyhedron with $n$ quadrangular faces abd vertices of order no greater than four (true in all cases for $n<10$) corresponds with one having $2n$ pentagonal faces.

When seven is an unlucky number

Consider the figure below, in which the cells of a square grid are subdivided by the blue lines to form trapezoidal and triangular pieces. A trapezoid plus the triangle sharing a common edge of the square grid form a pentagon.

On the left we extract six cells in the form of an unfolded cube. When these are wrapped up to reconstitute the cube, the dividing lines within the squares form a graph corresponding to the regular dodecahedron, which can be adjusted by moving the blue vertices outwards to become conspherical with the cube vertices.

On the right there are nine divided square cells, which can be wrapped up by tucking the left square into the right corner, folding the top and bottom triplets together, and deforming the squares into rhombi or kites to fit the occluded edges together. This gives the Herschel enneahedron (J??ErYsw?w? in graph6 encoding), a polyhedron having $D_{3h}$ symmetry; the blue lines then graphically define one of two mirror-image pentagonal 18-hedra with $D_3$ symmetry (the symmetry is reduced because the division of the squares is not fully symmetric). Like the cube and regular dodecahedron, the enneahedron and mirror-image pair of 18-hedra are graphichally unique for the number and polygonal type of faces.

We then find that there is no way to wrap a set of exactly seven contiguous squares into a polyhedron, even allowing for deformation of the faces to fit edges together. So the non-existences of a quadrangular heptahedron and a pentagonal tetrakaidecahedron fall together.

Answer 16

Kind of similar to your example of the triangle whose sides are consecutive in Pascal's triangle: I asked a question about Pythagorean right triangles with repdigit areas, and it seems (but is still unproven as far as I know) that the triangle with sides $\{693, 1924, 2045\}$ and area $666666$ is the only one.

Answer 17

"Let be $U$ non-empty simply connected open subset of $\mathbb C$, then there is a biholomorphic mapping from $U$ onto the open unit disk." has only a counterexample: $U = \mathbb C$.

Answer 18

Let be $\gamma$ a closed curve surrounding 0: $$\int_\gamma z^n\,dz = 0$$ for all $n\in\mathbb Z,\ n\ne -1$.

Answer 19

Given any additive sequence $\langle a_n \rangle$ of real numbers with first term $1,[a_0 = 1, a_1 = b, \, \text{and for all} \,\, n > 1, a_n = a_{n-1} + a_{n-2}]$, there exists no real number $b$ as second term, such that the limit of the ratio $a_n / a_{n-1}$ as $n \to \infty$, fails to approach golden number $\phi = \dfrac{(1 + \sqrt 5)}{2}$.

The single counterexample is $b = -\dfrac{1}{\phi} = \dfrac{(1 - \sqrt 5)}{2}$.

SOURCE: Michael W. Ecker, Generalized Additive Sequences from Fibonacci Numbers, Mathematics and Computer Education, Winter 2011-2012, Vol. 46, No. 1, p. 41-52.

Answer 20

There is no vector space $R^n$ with $n\ne 3$ where it is possible to define a not trivial cross product, with the only exception of $n=7$, where we have a family of 480 different cross products.

Answer 21

The line graph $L(G)$ of a graph $G$ is defined as $V(L(G)) = E(G)$, two vertices of $L(G)$ are adjacent if and only if their corresponding edges share a common vertex.

The structure of a connected graph $G$ can be recovered completely from its line graph with one exception. Meaning if $G \not\simeq H$ then $L(G) \not\simeq L(H)$, except for $L(K_3) \simeq L(K_{1,3})$.

Answer 22

$$ 3^3 + 4^4 + 3^3 + 5^5 = 3435 $$

is the only positive integer with this property other than trivial $1^1 = 1$.

Answer 23

Let $\ell^{p}$ be the space of complex sequences that are $p-$summable, where $p \in [1,\infty)$. Then non of them is a Hilbert space.

There is unique counterexample which is when $p=2$.

see:$\ell_p$ is Hilbert if and only if $p=2$

Answer 24

$$\int x^ndx=\frac{x^{n+1}}{n+1}+C$$

Alternate formulation with less obvious exception:

$$(n+1)\int x^ndx=x^{n+1}+C^{\prime}$$

Answer 25

A tournament is a complete graph with $n$ players in which each edge has been given an orientation. (The name comes from the idea that a tournament encodes the results of a round-robin contest, with an edge from $u$ to $v$ indicating that $u$ won her game against $v$.)

A champion in a tournament $T$ (sometimes called a king) is a node $c$ of $T$ where there is a directed path of length at most 2 from $c$ to each player in $T$. A classic theorem on tournaments states that every nonempty tournament has at least one champion.

The question then arises: given $n \ge 1$ and $1 \le k \le n$, is there an $n$-player tournament with exactly $k$ champions? The answer is yes, except if $k = 2$ or $n = k = 4$.

(This technically has more that one counterexample, but the $(4, 4)$ case definitely stands out!)

Answer 26

"Let $G$ be a finite group and $p\ne q$ prime integers. If $G$ does not have any elements of order $pq$, then the Sylow $p$-subgroups or the Sylow $q$-subgroups of $G$ are abelian." has only a counterexample: the monster group.

Source:

https://www.quora.com/What-are-some-weird-theorems-in-mathematics/answer/Colin-Reid-5

https://agag-malle.math.rptu.de/~malle/download/elemords.pdf

Answer 27

Let $p(z) \ = \ z^n + p_{n-1}z^{n-1} + \ldots + p_1z + p_0$ be a complex polynomial. Then its polar derivative $q$ with respect to the pole $\alpha \ \in \ \mathbb{C}$ is defined by $$ q(z) \ = \ p(z) - \frac{1}{n}(z-\alpha)p´(z) \, , $$ where $p´(z)$ is its usual derivative. The claim that $q(z)$ is of degree $n-1$ has only one counterexample $\alpha \, = \, c$, where $c$ is the centroid of the zeros of $p$.

This is a direct consequence of the difference of the sum of the zeros and critical points (zeros of the derivative) always being equal to $c$ and the only reason that certain matrices I am particularly interested in exist at all.

In the case $\alpha \, = \, c$, the degree of $q$ is at most $n-2$, but if the coefficients of $p$ fulfill certain conditions, it will be lower still. The extreme case is $p(z) \, = \, (z-k)^n$, where $k$ is a fixed complex number. Then $k \, = \, c$ and $q$ with respect to $c$ vanishes.

Answer 28

Claim: Any orientable 3-manifold $M$ is prime if and only if it is irreducible.

This statement holds for all orientable 3-manifolds except for $S^1 \times S^2$, see Proposition 9.2.14. in An Introduction to Geometric Topology.

Definitions

A connected, oriented 3-manifold $M$ is called prime if whenever $M$ can be expressed as a connected sum
$$ M = M_1 \# M_2 $$ it follows that either $M_1$ or $M_2$ is the 3-sphere $S^3$.

A connected, oriented 3-manifold $M$ is called irreducible if every smoothly embedded 2-sphere bounds a 3-ball.

$S^1\times S^2$ is prime but not irreducible

• It is not irreducible since the embedded spheres of the form $\{ pt \} \times S^2$ do not bound 3-balls, since the complement of such a sphere is $S^2 \times I$.
• To see why it is prime, consider the following argument (taken from here, page 280): Let $S\subset M$ be a separating sphere (the complement consists of two components). If $S$ bounds a ball the claim follows. Let $M$ and $N$ be the two connected manifolds whose union is the complement of $S$. Since $\mathbb{Z}\cong \pi_1(S^1\times S^2)=\pi_1(M) \star \pi_1(N)$, either $M$ or $N$ must be simply-connected. Without loss of generality assume $M$ is simply-connected. Then $M$ lifts to a copy $M'$ in the universal cover of $S^1 \times S^2$, which is homeomorphic to $\mathbb{R}\times S^2$. The latter manifold can be identified with $\mathbb{R}^3\setminus 0$. Using Alexander's theorem, stating that every embedded sphere in $\mathbb{R}^3$ bounds a ball, we obtain that $M'$, and therefore $M$, must be a ball.

Answer 29

Claim: There is no Pythagorean triple whose numbers are consecutive.

The one (well-known) counterexample is $3^2+4^2=5^2$.

Proof (easy but arguably not trivial):

If$$a^2+(a+1)^2=(a+2)^2$$then$$2a^2+2a+1=a^2+4a+4$$$$a^2=2a+3$$and$$a=3$$

Answer 30

The triangular graph of order $n$ is determined by its spectrum for all $n \neq 8$, for which it is cospectral to three other graphs known as the Chang graphs.

I believe this fact traces back to the existence of the exceptional root system $E_8$.

Answer 31

Let $V$ be a finite dimensional vector space over an arbitrary field $K$, and let $Q$ be a quadratic form on $V$ whose associated bilinear form $B(x,y)=Q(x+y)-Q(x)-Q(y)$ is nondegenerate. Then the Cartan–Dieudonne theorem says that the orthogonal group $\operatorname O_Q(V)$ is generated by its reflections.

The only counterexample, up to isometry, is $$\begin{array}{c} K=\mathbb F_2\\ V=(\mathbb F_2)^4\\ Q(xe_1+ye_2+ze_3+we_4)=xy+zw. \end{array}$$

This is from Larry Grove's Classical groups and geometric algebra, Theorem 14.16.

Answer 32

This answer probably isn't quite what you had in mind, but it might arguably fit the criteria of your question:

The image of any non-constant entire function is the entire complex plane.

By Picard's little theorem, there can be a single complex number that is not in the image of the function (e.g. $0$ for the function $\exp(z)$).

Here the single counterexample is not a function, but an element of the set $\mathbb{C}$, the putative image of the function. But it isn't a fixed complex number; it depends on the function in question.