Is there a deep reason why $(3, 4, 5)$ is pythagorean? [closed]

Put $a = n$, $b = n + r$, $c = n + 2r$. Simplify $c^2 = a^2 + b^2$ to get: $$ (n + r)(n - 3r) = 0 $$

Either $n = -r$, but this means $b = 0$. Or $n = 3r$, which gives: $$ a = 3r,\ b = 4r,\ c = 5r $$

Therefore, $(3, 4, 5)$ (and its multiples) is the only arithmetic progression that is also a Pythagorean triple.

Actually these are the only 3 natural consecutive numbers that match equation.

We are looking for solution for this equation:

$$\begin{align}a^2+(a+1)^2&=(a+2)^2\\a^2-2a-3&=0\end{align}$$ And the only solutions are $a_1=3, a_2=-1$.

And I don't think there is any meaning in these numbers.

$(3,4,5)$ is Phythagorean because $5$ is a prime of the form $4k+1$. Some known facts:

• Every prime $p$ of the from $4k+1$ can be rewritten as a sum of squares of two distinct positive integers:

$$\forall k \in \mathbb{Z}_{+}, p = 4k+1\text{ prime} \implies \exists \alpha, \beta \in \mathbb{Z}_{+} \text{ s.t. } \alpha \neq \beta \wedge p = \alpha^2 + \beta^2$$

• Every number $n$ that can be written as a sum of squares of two distinct positive integers is part of a Pyhthagorean triplet because of an algebraic identity:

$$n = (\alpha^2+\beta^2) \implies n^2 = (\alpha^2+\beta^2)^2 = (\alpha^2-\beta^2)^2 + (2\alpha\beta)^2$$

• Every Phythagorean triplet $(a,b,c)$ has a parametrization of the form:

$$a^2 + b^2 = c^2 \implies \begin{cases}a = (\alpha^2-\beta^2)\mu\\b = 2\alpha\beta\mu\\c = (\alpha^2 + \beta^2)\mu\end{cases}\quad\quad\text{up to order of }a, b$$

• When $a, b$ is relative prime to each other, we can set $\mu$ above to 1.

Take $5 = 2^2+1^2$ as an example, we get:

$$\begin{cases}a = 2^2-1^2 = 3\\b = 2\cdot 2 \cdot 1 = 4\\c = 2^2 + 1^2 = 5\end{cases} \quad\quad\text{is a Pythagorean triplet}$$

$c = 5$ is the smallest example of such Pythagorean triplet. Since there are only 4 numbers smaller than 5, it is just a coincidence that $(3,4,5)$ are successive integers.

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.

Well, you can characterize all triples by

$a=m^2-n^2, b=2mn, c=m^2+n^2$ with $m$ and $n$ co-prime.

If you choose the smallest such pair, $m=2, n=1$, you get $3,4,$ and $5$. So, in a sense, it's the simplest tripple you can construct.