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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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    $\begingroup$ Hmm, $3^2+4^5=5^2$ and $3^3+4^3+5^3=6^3$, but $3^4+4^4+5^4+6^4\ne7^4$, and on the other side $3^1\ne4^1$... $\endgroup$
    – user856
    Commented Jun 16, 2013 at 0:15
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    $\begingroup$ Nevertheless, I vote against closing this question. If you would like to add a close vote, please post a comment cancelling my anti-vote instead. $\endgroup$
    – user856
    Commented Jun 16, 2013 at 0:15
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    $\begingroup$ I think this question is too vague, and the asker doesn't really quantify what they mean by "deep reason". The top two answers are actually answering the different but related questions, what are the possible pythagorean triples where the three numbers are consecutive (or more generally are in an arithmetic progression. $\endgroup$ Commented Jun 16, 2013 at 14:17
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    $\begingroup$ My hunch (which, admittedly, is just that) is that the arithmetic progression aspect is probably just a red herring, and the real question behind this is closer to "Why should there be such a small Pythagorean triple as (3,4,5)?", or even "Why are there (integer) Pythagorean triples at all?" After all, there aren't any for higher exponents. $\endgroup$ Commented Jun 16, 2013 at 20:03
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    $\begingroup$ $$\begin{gathered} {\left( {n - 1} \right)^2} + {n^2} = {\left( {n + 1} \right)^2} \\ {n^2} - 4n = 0\\ n\left( {n - 4} \right) = 0 \\ n = 0{\text{ or }}4 \\ (-1,0,1) \mbox{and} (3,4,5)\\ \mbox{Do you consider this as "DEEP"?}\\ \end{gathered} $$ $\endgroup$ Commented Jun 21, 2013 at 3:44

5 Answers 5

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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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    $\begingroup$ I don't really see how this answers the question - he asked for a deep reason, whereas you gave an algebraic reason. $\endgroup$
    – Daniel
    Commented Jun 17, 2013 at 18:19
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    $\begingroup$ @Daniel What's the definition of a "deep" reason? $\endgroup$ Commented Jun 17, 2013 at 19:06
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    $\begingroup$ Why did you answer a question asking for a deep reason if you didn't understand what a deep reason is? $\endgroup$
    – Daniel
    Commented Jun 17, 2013 at 22:53
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    $\begingroup$ @Daniel My response to you was a rhetorical question if you couldn't tell. The word "deep" is subjective. The OP asked for an explanation as to why an arithmetic progression is a Pythagorean triple. I feel that my answer addresses this. All of the other answers follow (what you call) an algebraic approach too. $\endgroup$ Commented Jun 17, 2013 at 23:26
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    $\begingroup$ Well this is really just a pointless semantic debate about the meaning of the phrase 'deep reason.' It's definitely a subjective phrase so I suppose that saying that your answer didn't answer the question was wrong. Apologies :). $\endgroup$
    – Daniel
    Commented Jun 17, 2013 at 23:40
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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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  • $\begingroup$ Not all triples. $a=k(m^2-n^2), b=2mnk, c=k(m^2+n^2)$ does characterize all triples though. $\endgroup$
    – user26486
    Commented Jan 26, 2015 at 21:51

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