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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 $3^3+4^3+5^3$.

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5 Answers

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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.

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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.

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$(3,4,5)$ 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.

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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.

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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.

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