TITLE: Is there a deep reason why $(3, 4, 5)$ is pythagorean?
QUESTION [21 upvotes]: The triple $(3, 4, 5)$ is a pythagorean triple - it satisfies $a^2 + b^2 = c^2$ and, equivalently, its components are the lengths of the sides of a right triangle in the Euclidean plane.
But of course, the first thing anybody notices is that the triple $(3, 4, 5)$ also happens to be an arithmetical succession of small numbers.
Is there a deep reason why choosing these three successive numbers just so happens to yield a pythagorean triple?
To anyone who feels the question is silly: consider $3^3+4^3+5^3$.

REPLY [7 votes]: Pythagorean triples with two consecutive numbers
Actually there are infinite Pythagorean triples in which the two highest numbers are consecutive, with the condition that the sum of these numbers is a square.  The proof is really straightforward. Let $a$ be a natural number, the difference between the squares of $a$ and $a+1$ is
$$(a+1)^2 - a^2 = 2a + 1 = a + (a+1).$$
$a$ and $a+1$ constitutes a Pythagorean triple if $2a+1$ is also a square.  Of course the lowest number must be odd and indeed all odd numbers, except 1, can be used to construct such triples. Examples are $(3,4,5)$, $(5,12,13)$, $(7,24,25)$, $(9,40,41)$, $(11,60,61)$, $(13,84,85)$, etc.
Pythagorean triple with three consecutive numbers
If you in addition want that the lowest number precedes the central one you have to do other calculations. Let $2n+1$ be the lowest number, with $n$ natural, its square is the sum of the highest ones:
$$(2n + 1)^{2} = 4n^{2} + 4n + 1 = 2n(n + 1) + (2n(n + 1) + 1)$$
Thus, the general form of these triples is $(2n+1, 2n(n+1), 2n(n+1) +1)$.  If $2n+1$ precedes $2n(n+1)$ the following equation holds
$$(2n + 1) + 1 = 2n(n + 1) \iff 2(n+1) = 2n(n+1)$$
from which $n = 1$ and the wanted triple is $(3,4,5)$.
Hence, there is really nothing special in a Pythagorean triple with two consecutive numbers, $(3,4,5)$ is just the only triple with all three numbers consecutive.