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Here are two classical results which depend on whether a parameter is 2 or 3:

  • It is possible to bisect an arbitrary angle with ruler and compass, but impossible to trisect it.

  • While there are infinitely many Pythagorean triples, i.e. integer solutions to $x^2+y^2=z^2$, there are no non-trivial integer solutions to $x^3+y^3=z^3$.

There are several other instances where the dividing line seems to be between 2 and 3:

  • A 2-regular tree is countable, a 3-regular tree is uncountable.

  • 2SAT is solvable in polynomial time, 3SAT is NP-complete.

  • A random walk on $\mathbf Z^2$ is recurrent, while a random walk on $\mathbf Z^3$ is transient.

What other examples can you think of?

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closed as not a real question by Andres Caicedo, Felipe Voloch, Charles Siegel, darij grinberg, Zev Chonoles Apr 18 '11 at 16:02

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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All the prime numbers less than or equal to 2 are even, and all the prime numbers greater than or equal to 3 are odd :) –  Will Merry Apr 18 '11 at 15:18
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I'm sorry. I may just be in a poor mood (those who follow other parts of the internet math Q&A community will catch an allusion here) but at the moment your question strikes me as somewhat superficial: $2$ is not equal to $3$, so there are going to be a lot of instances where changing $2$ to $3$ makes a big difference. But perhaps there is a good question lurking in here somewhere, something like: what common explanations can be given for these examples? It might be worth thinking about how to rephrase it. –  Pete L. Clark Apr 18 '11 at 15:19
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I don't think it's just you being in a bad mood, Pete. It would at least help if $3$ was replaced by $n\geq 3$ in the places where it can be, I guess. –  Tara Brough Apr 18 '11 at 15:25
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While others have essentially said it already I need to say it too: this is so so broad and vague; hundreds of answers could be given, and I have the strong feeling that they will be given and the couple of interesting ones will be hard to find in this flood, while the question will be on the front page for way too long. –  quid Apr 18 '11 at 15:35
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A correct answer to the question in the title would be 1 I believe. –  Roland Bacher Apr 18 '11 at 15:39
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8 Answers 8

There are infinitely many regular polytopes in $\mathbb R^2$ and only five in $\mathbb R^3$.

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$SL_2(\mathbb{Z})$ is an amalgam whereas $SL_3(\mathbb{Z})$ is not.

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Autonomous systems of ODEs produce simple dynamics in two dimensions, but complex dynamics in three or more. This is directly related to the fact that curves in three or more dimensions can pass each other without crossing.

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The permutation group of two elements is abelian, the permutation group of three elements is not. There are thus non-galoisian number fields of degree $3$.

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More examples are given as answers to a similar question about problems NP-hard in $\mathbb R^3$ but not in $\mathbb R^2$:

  • Set-cover by half-spaces.
  • Finding a shortest path between two points among polygonal obstacles.
  • Determining whether a non-convex polygon/polyhedron can be triangulated without Steiner points.
  • Realizability problem for $d$-dimensional polytopes is a candidate ($d \leq 3$ vs $d \geq 4$).
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In 2-dimensional Euclidean space every two lines intersect (maybe in the infinite), in 3-dimensional Euclidean space there are skew (?) lines.

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$\mathbb R^3$ is much more rigid than $\mathbb R^2$ when considering conformality: Conformal transformations of $\mathbb R^2$ do not form a finite-dimensional Lie-group.

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Elements of order $2$ in a group are the only non-trivial elements which their own inverse.

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