# Talk:Pythagorean triple

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## A New Property -- Can Anyone Find it in the Literature?

For primitive PT's: c2+2ab and c2-2ab are both odd squares.

The proof is simple: c2+2ab=a2+b2+2ab=(a+b)2. The latter is an odd square because exactly one of a, b is odd. Similarly for c2-2ab.

I don't see this in my copy of Sierpinski's "Pythagorean Triangles"... has anybody seen it somewhere? If so, is it worth adding to "Elementary Properties..."? (And if not, this has to be the most trivial "original research" you can imagine.) CirclePi314 (talk) 09:20, 12 February 2015 (UTC)

This article tends to be a magnet for things that match the description in your final parenthetical. --JBL (talk) 16:29, 12 February 2015 (UTC)

## Another Property -- Can Anyone Find it in the Literature?

Some triplet comes also from:

${\displaystyle A^{2}=(C+B)/(C-B)}$

A=3, B=4, C=5 is the most famous. — Preceding unsigned comment added by 94.81.217.96 (talk) 13:15, 19 May 2016 (UTC)

Well, we know that ${\displaystyle A^{2}=C^{2}-B^{2}=(C+B)(C-B)}$, so ${\displaystyle A^{2}=(C+B)/(C-B)}$ if and only if ${\displaystyle C-B=\pm 1}$. For some Pythagorean triplets it is the case that ${\displaystyle C-B=1}$, but there are also many Pythagorean triplets for which ${\displaystyle C-B\neq \pm 1}$. 𝕃eegrc (talk) 14:13, 19 May 2016 (UTC)

## Dubious passage on Platonic sequences

The following was put into the article's section on Platonic sequences on 28 June 2007 by someone who is no longer on Wikipedia:

It can be shown that all Pythagorean triples are derivatives of the basic Platonic sequence (xyz) = p, (p2 − 1)/2 and (p2 + 1)/2 by allowing p to take non-integer rational values. If p is replaced with the rational fraction m/n in the sequence, the 'standard' triple generator 2mn, m2 n2 and m2 + n2 results. It follows that every triple has a corresponding rational p value which can be used to generate a similar triangle (one with the same three angles and with rational sides in the same proportions as the original). For example, the Platonic equivalent of (6, 8, 10) is (3/2; 2, 5/2). The Platonic sequence itself can be derived by following the steps for 'splitting the square' described in Diophantus II.VIII.

This seems wrong to me. First, as far as I can see, the Platonic sequence p, (p2 − 1)/2 and (p2 + 1)/2 does not give rise to (3/2, 2, 5/2). Second, as far as I can see it is impossible to find any rational p for which this formula gives something that scales to (6, 8, 10). Third, I think the same is true for the different Platonic sequence formula given in the article Diophantus II.VIII.

I conclude from this that in fact the Platonic sequence can generate all primitive triples but not all triples. This feeling is reinforced by the observation in the above passage that the Platonic sequence can be used to derive the 'standard' formula, which generates all primitive triples but which does not generate all triples.

Without objection, I'm going to replace the above passage with the following:

''It can be shown that all primitive Pythagorean triples can be derived from the basic Platonic sequence (xyz) = p, (p2 − 1)/2 and (p2 + 1)/2 by allowing p to take non-integer rational values. If p is replaced with the rational fraction m/n in the sequence, the 'standard' primitive triple generator 2mn, m2 n2 and m2 + n2 results. It follows that every primitive triple has a corresponding rational p value which can be used to generate a similar triangle (one with the same three angles and with rational sides in the same proportions as the original). For example, the Platonic equivalent of (56, 33, 65) is generated by p = m/n = 7/4 as (p, p2 –1, p2+1) = (56/32, 33/32, 65/32). The Platonic sequence itself can be derived by following the steps for 'splitting the square' described in Diophantus II.VIII.

Loraof (talk) 17:58, 8 April 2015 (UTC)

I don't understand your objection. If I can make something similar to every primitive triple then I can make something similar to every triple (since every triple is similar to a primitive one). Though the example is confused, since to get the triple (3/2, 2, 5/2) one should put in p = 2 (an integer). --JBL (talk) 18:32, 8 April 2015 (UTC)
Thanks. First, I found (and still find) the phrase "triples are derivatives of ..." confusing. From your comments I can see that this was intended to mean with unrestricted scaling. Second, I was misled by the statement that the Platonic sequence, previously defined as p, (p^2-1)/2, (p^2+1)/2, is exemplified by (3/2, 2, 5/2), which says that p=3/2 and 2=((3/2)^2-1)/2, which is not true. It should be (2, 3/2, 5/2). I'm going to clarify the derivatives wording, and replace the example with one that is correctly sequenced and has a fractional p. Loraof (talk) 19:16, 8 April 2015 (UTC)
Yes, you're certainly right that it was not clearly written at all. Your edit was a big improvement; in a second I will tweak it a bit more. --JBL (talk) 21:30, 8 April 2015 (UTC)

## Pairwise coprime

Since for Pythagorean primitive triples (a,b,c) we have gcd(a,b) = gcd(a,c) = gcd(b,c) = 1. I propose to give this as a property in the article. — Preceding unsigned comment added by 147.215.1.189 (talk) 12:43, 6 October 2015 (UTC)

## Rightarrow versus implies

In Google Chrome on my laptop running Windows Vista, in the Points on a unit circle section, the \implies produces an arrow which is broken in the middle, but the \Rightarrow looks fine. This may be why the IP 128.84.127.40 changed it. — Anita5192 (talk) 21:14, 29 January 2016 (UTC)

Would it be better just to replace the arrow with the words "and so" or similar? --JBL (talk) 21:20, 29 January 2016 (UTC)
I think it would be equally readable with "and so," or "and," or ${\displaystyle \Rightarrow }$. I do not understand why ${\displaystyle \implies }$ looks strange on my computer, or what is wrong with using ${\displaystyle \Rightarrow }$. I was hoping someone else would see the same problem with ${\displaystyle \implies }$, and understand why I mentioned this. — Anita5192 (talk) 21:39, 29 January 2016 (UTC)
I also see a funny kink in the symbols \implies and \Longrightarrow. (I am using Firefox on a unix machine.) --JBL (talk) 22:17, 29 January 2016 (UTC)
My general preference in situations like this is to spell it out in words. I think doing it that way makes it less WP:TECHNICAL. —David Eppstein (talk) 23:19, 29 January 2016 (UTC)
Done! Thank you both for your input. I have not heard from D.Lazard, but I hope he approves. — Anita5192 (talk) 02:47, 30 January 2016 (UTC)

Since I am an author of the article referenced in that page, I don't know whether I should add the link just myself (which seems harmless, since it is another Wikipedia page, and obviously it is related to the current page). Oliver Kullmann (talk) 16:34, 5 June 2016 (UTC)

Done D.Lazard (talk) 07:52, 6 October 2016 (UTC)

## Euclid's formula when a is even

In the text it says that "A proof of the necessity that ${\displaystyle a,b,c}$ be expressed by Euclid's formula for any primitive Pythagorean triple is as follows". However, Euclid's formula only gives primitive Pythagorean triples with ${\displaystyle b}$ even. It is for instance obvious that we cannot find m and n such that ${\displaystyle m^{2}-n^{2}=4}$, ${\displaystyle 2mn=3}$, ${\displaystyle m^{2}+n^{2}=5}$. As pointed out earlier, if ${\displaystyle m-n}$ is even and ${\displaystyle \gcd(m,n)=1}$, Euclid's formula gives ${\displaystyle (2a,2b,2c)}$ where ${\displaystyle (a,b,c)}$ is a PPT. For instance ${\displaystyle m=3}$ and ${\displaystyle n=1}$ gives ${\displaystyle (8,6,10)}$.

So a few places, primitive Pythagorean triple should be replaced by primitive Pythagorean triple ${\displaystyle (a,b,c)}$ where ${\displaystyle b}$ is even. — Preceding unsigned comment added by HelmerAslaksen (talkcontribs) 19:44, 6 September 2016 (UTC)

Euclid's formula produces the triple (3, 4, 5) with m = 2, n = 1. It also produces every other primitive Pythagorean triple. It also produces a limited family of non-primitive triples, but that is irrelevant to the claim in question. --JBL (talk) 21:06, 6 September 2016 (UTC)
Dear Joel, I'm afraid you didn't understand my post, or we disagree about what we mean by Euclid's formula. My point is that what I call Euclid's formula, ${\displaystyle (m^{2}-n^{2},2mn,m^{2}+n^{2})}$, will only produce triples ${\displaystyle (a,b,c)}$ with ${\displaystyle b}$ even, and not for instance ${\displaystyle (4,3,5)}$. If you want triples with ${\displaystyle b}$ odd, you must reorder the formula, or include the values of ${\displaystyle m}$ and ${\displaystyle n}$ with ${\displaystyle m-n}$ even, and then divide by 2. Not sure what you mean by "limited family". Half the time (${\displaystyle m-n}$ even) you will get a non-primitive triple.

HelmerAslaksen (talk) 00:34, 6 October 2016 (UTC)

I have fixed this, but it would be better to say (where?) that two triples that differ by the exchange of a and b are considered as equal (because of the commutativity of addition) This was implicitly supposed in the article and is the cause of the confusion. D.Lazard (talk) 07:44, 6 October 2016 (UTC)
The reason why I started this discussion was because I was lecturing about this to a class, and I suddenly realized that it was more difficult that I had first thought. First of all, I believe that we should make our assumptions clear. Otherwise, both we and our students, may suddenly find ourselves getting confused. Secondly, I feel that this explains the ${\displaystyle m-n}$ condition. The ones with ${\displaystyle m-n}$ odd are the primitives with ${\displaystyle b}$ even, and the ones with ${\displaystyle m-n}$ even are the triples of the form ${\displaystyle (2a,2b,2c)}$ where ${\displaystyle (a,b,c)}$ is a primitive triple with ${\displaystyle b}$ odd. It was only when I saw that, that I felt that I really understood this. — Preceding unsigned comment added by HelmerAslaksen (talkcontribs) 13:28, 6 October 2016 (UTC)

Take a look at Pythagorean triple#Interpretation of parameters in Euclid's formula. There we see that n/m is the lowest terms representation of tan(θ/2) in the Pythagorean triangle where θ is the angle opposite the side of length a. The reason that sometimes we have to divide by two to make the Euclid-generated Pythagorean triple be primitive is that the focus on tan(θ/2) is in some sense inappropriate. If we instead look at tan(θ) = a/b in lowest terms (or look at sin(θ) = a/c or cos(θ) = b/c in lowest terms) we would have two-thirds of our primitive Pythagorean triple immediately, never having to divide by 2. Fortunately, the double-angle formula for a tangent is easy enough to analyze; tan(θ) = 2 tan(θ/2) / (1 - tan2(θ/2)) = 2mn / (m2 - n2) will be in lowest terms exactly when n and m are relatively prime and not both odd. If they are relatively prime but both odd then the lowest terms representation for tan(θ) requires that both the numerator 2mn and denominator m2 - n2 be halved. 𝕃eegrc (talk) 16:29, 6 October 2016 (UTC)

I have added a section "A variant". I have written it in order that both parameterizations of triples (Euclid's and its variant) parameterize the primitive triples with b even. But this could be formulated without exchanging a and b by saying that every primitive triple may be obtained exactly once by choosing m and n coprime and dividing by two if m and n are both odd. D.Lazard (talk) 17:51, 6 October 2016 (UTC)

When m and n are coprime but both odd then (m′, n′) = ((m+n)/2, (mn)/2) will be coprime, not both odd, and will generate the primitive Pythagorean triple associated with (m, n), but with a and b reversed. This follows, in part, from the fact that the two non-right angles in a Pythagorean triangle sum to π/2, so their half angles sum to π/4, and the formula tan(π/4 - θ/2) = (1 − tan(θ/2)) / (1 + tan(θ/2)). 𝕃eegrc (talk) 15:53, 7 October 2016 (UTC)