What are some interesting sole exceptions or counterexamples?

Question

Many theorems assert that a particular property holds for all objects in a class except those in a given list of exceptions. Examples of rules that admit precisely one exception include:

• All primes are odd, except for $2$
• All automorphisms of $S_n$ are inner for all $n$ except $6$
• All simple Lie algebras have abelian outer automorphism group except for $D_4$ ($\text{Out}(D_4) \cong S_3$, which leads to the exceptional phenomenon of triality)

What are some other interesting examples of results that admit (essentially) one exception?

Edit (modified from a comment below): Read strictly this question is subordinate to the one someone suggested it duplicated, but that question asks (more or less) about classifications of exceptional objects more generally, and not specifically about the case of a single exception. Partly because of this, very few of the answers to that question are admissible answers to this one (and two of them are actually already given in the question here).

Answer 1

The smooth structure on $\mathbb{R}^n$ is unique up to diffeomorphism, except if $n = 4$.

Answer 2

If the $n$th Fibonacci number is prime then $n$ is prime, except that $F_n=3$ when $n=4$.

Answer 3

The group of units of $\mathbb Z/p^n \mathbb Z$ is cyclic for every prime power $p^n$, except when $p=2$; then it's cyclic only for $n=1,2$.

Answer 4

A generalization of the Four Color Theorem says that the chromatic number of a closed surface with Euler characteristic $\chi$ (the number of colors needed to color any map on the surface) is bounded above (sharply) by $$\left\lfloor \frac{7 + \sqrt{49 - 24 \chi}}{2} \right\rfloor,$$ except for the Klein bottle, which has Euler Characteristic zero, so that the above formula gives a count of $7$ colors, but which only requires $6$. (See Ringel, G. and Youngs, J. W. T. Solution of the Heawood Map-Coloring Problem. Proc. Nat. Acad. Sci. USA 60, 438-445, 1968.)

(This is borrowed from this answer to the question JimmyK4542 references in the comments above.)

Answer 5

Here are a couple involving zero, that students often forget to their chagrin:

The square of every real number is positive, except for $0$.

Every real number has a multiplicative inverse, except for $0$.

Answer 6

A lot of answers to this question can be given by taking unique answers and negating them. The one that came to mind just now was "the number of critical points of a Morse function on an $n$-dimensional manifold is not equal to 2 unless the manifold is an $n$-sphere." We also have things like "a division algebra over the reals of any finite dimension except 8 is associative."

Answer 7

The duals of $L^p$ spaces: $$1\le p<\infty,\ \frac1p+\frac1q = 1\implies (L^p)^*=L^q,$$ but $$(L^\infty)^*\ne L^1.$$

Answer 8

All spheres $S^n$ are simply-connected except $S^1$.

Answer 9

$$\forall x,y \in \mathbb{Z_+}$$ $$|y^3-x^2| \ne 2 $$ except for $x = 5$ and $y=3$

Ie 26 is the only number between a square and a cube.

Answer 10

Zsigmondy's theorem.

If $a,b\in\mathbb Z, n\in\mathbb N_{\ge 2}, (a,b)=1, a>b$, then $$\exists p\in\mathbb P( \forall k\in\mathbb N\cap[1;n)(p\mid a^n+b^n\wedge p\not\mid a^k+b^k))$$

The only exception to this is $(a,b,n)=(2,1,3)$.

---

The following variation of it

If $a,b\in\mathbb Z, n\in\mathbb N_{\ge 2}, (a,b)=1, a>b$, then $$\exists p\in\mathbb P( \forall k\in\mathbb N\cap [1;n)(p\mid a^n-b^n\wedge p\not\mid a^k-b^k))$$

has more than one exception, namely $(a,b,n)=(2,1,6)$ and any pair $(a,b,n)=(k,l,2)$ with $k+l=2^m$ for some $m\in\mathbb N$.

Answer 11

Just an observation from $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$:

For any prime $p$ such that $p\neq 2$, $$\sqrt{p+\sqrt{p+\sqrt{p+\cdots}}}<p$$ and $$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}=2.$$

Answer 12

For any prime $n$ except $2$, $x^n+y^n=z^n$ has no solution in nonzero integers.

Answer 13

$x^a-y^b=1$ has no solution in prime numbers, except $3^2-2^3$.

Answer 14

Fermat's last theorem: There are no solutions in $\mathbb{Z}$ in the equation $a^n+b^n=c^n$, except for $n=2$

$0$ properties:

The only number that does not have a multiplicative inverse is $0$.

The only number that satisfies $-x=x$ is $0$.

$\frac xx=1, x \not \in \{0\}$

$\frac 0x=0, x \not \in \{0\}$

$x^x\in \mathbb{C}, x \not \in \{0\}$

Answer 15

Here is one I found. The sum of two complex sinusoids with the same frequency is another sinusoid of that frequency, with 1 exception:

For amplitudes $A_1,A_2 \in \mathbb C$ and phases $\varphi_1, \varphi_2 \in \mathbb R$ and frequency $f \in \mathbb R$:

If $f_1(x) = A_1\cos(2\pi f~x + \phi_1)$ and $f_2(x) = A_2 \cos(2\pi f ~x + \phi_2)$

Then there is an amplitude $A_3 \in \mathbb C$ and phase $\varphi_3 \in \mathbb R$ such that

$f_1(x) + f_2(x) = A_3\cos(2\pi f~x + \varphi_3)$

with the exception:

$f_1(x) + f_2(x) = \cos(x) \pm i~\sin(x)$

...and all transforms of the above exception by moving/scaling the coordinate system.

Answer 16

All compact orientable two dimensional surfaces admit a metric with a constant non-positive curvature, except the sphere $S^2$.

Answer 17

For all $n$, the braid group on $n$-strings $B_n(M)$ of a closed orientable surface $M$ is torsion free unless $M=S^2$, the $2$-sphere.

The reason the same proof for other surfaces can't be used for the sphere is that the sphere is the only closed surface with non-trivial higher homotopy (because all other surfaces have a contractible universal cover, but the sphere is its own universal cover). This means that the Fadell-Neuwirth fibration sequence

$$F_{m+r,n−r} S^2 \to F_{m,n} S^2 \to F_{m,r} S^2$$

(here $F_{m,n}M$ is the ordered configuration space of $m$ point in $M\setminus\{n\mbox{ points}\}$) induces a long exact sequence in homotopy which isn't bounded at $\pi_1$, but has non-trivial groups appearing for all higher $\pi_k$ and so importantly, $F_{n,0}(S^2)$ is not an Eilenberg-Maclane space. This is crucial in the proof for other surfaces because the fundamental group of a $K(G,1)$ with the homotopy type of a finite CW-complex is torsion-free by an old theorem of P.A. Smith.

The reason such groups actually do have torsion is a consequence of the well-known 'Dirac string trick'. Essentially, if you apply a full twist to the trivial braid, the new braid it is not trivial, but if you do two full twists, then you can isotope the braid to the trivial braid by 'pulling the strings around the sphere'

Answer 18

There are Graeco-Latin squares of all orders greater than two except of order six.

Answer 19

No $3$ consecutive odd integers are prime except $3,5,7$.