Talk:Pythagorean quadruple

I don't see the purpose of developing the case q=0. Setting any other parameter to zero also simplifies the formulae but does not give all values. &mdash; MFH:Talk 14:32, 20 April 2008 (UTC)
 * Neither do I. I'm removing it. Algebraist 16:33, 19 December 2008 (UTC)

2 Year old error?
I referred to the cited source (Carmichael) for some more information on this topic and I noticed that the 'a' equation has apparently been listed incorrectly for almost two years now. i'm going to correct it, but I thought I should check with anyone who cares to double check me. I saw it on p. 43 of the pdf version of Carmichael.

Whoa ... nevermind. The 'c' equation looks wrong too. They should be (from carmichael): t = m^2 + n^2 + p^2 + q^2 , x = m^2 − n^2 − p^2 + q^2 , y = 2mn − 2pq, z = 2mp + 2nq.

Can somebody with some math-mojo check this out? I'm afraid I'm wait out of my depth here. Davidyorke (talk) 10:33, 15 February 2009 (UTC)


 * I think it's just that the variables names for n and q are swapped. Daggerbox (talk) 23:45, 25 March 2011 (UTC)

Alternate parametrization
It was pointed out here http://math.stackexchange.com/questions/76892/pythagorean-quadruples that the parametrization given in the current version http://en.wikipedia.org/w/index.php?title=Pythagorean_quadruple&oldid=442018551#Alternate_parametrization of the article might be wrong. I have found another parametrization, but not the one given in article (or a corrected version of it). I am adding citation needed template at the moment.

Parametrization given in Andreescu et al: An Introduction to Diophantine Equations, Theorem 2.2.3, p.79, for solutions of $$x^2+y^2+z^2=t^2$$ is
 * $$x=\frac{l^2+m^2-n^2}n$$, $$y=2l$$, $$z=2m$$, $$t=\frac{l^2+m^2+n^2}n$$

where l, m are arbitrary positive integers and n is any divisor of $$l^2+m^2$$ less than $$\sqrt{l^2+m^2}$$



--Kompik (talk) 11:22, 29 October 2011 (UTC)

Characterization of "d"
What numbers are the diagonal d of a primitive Pythagorean quadruple? (The answer is simple of Pythagorean triple; an odd number which is the sum of two relatively prime squares.) Here. the answer would be any odd number which is the sum of three or four non-zero relatively prime squares. (The only problem might be c = 0 if the product m n p q is a square, but that can be resolved by permuting the parameters unless all of them are the same. It might be interesting, if we can find a source.  — Arthur Rubin  (talk) 05:55, 15 August 2014 (UTC)
 * It follows from inspection of that any odd number other than 1, 3, 5, 9, 11, 17, 29, 41 is the sum of 4 nonzero squares
 * $$3 = 1^2 + 1^2 + 1^2$$
 * $$9 = 2^2 + 2^2 + 1^2$$
 * $$11 = 3^2 + 1^2 + 1^2$$
 * $$17 = 3^2 + 2^2 + 2^2$$
 * $$29 = 4^2 + 3^2 + 2^2$$
 * $$41 = 6^2 + 2^2 + 1^2$$
 * Leaving only 1 and 5 as the only odd numbers which cannot be expressed as the sum of 3 or 4 non-zero squares. It follows that the only numbers which cannot be d are:
 * $$2^n, 5\cdot2^n$$
 * — Arthur Rubin (talk) 06:31, 15 August 2014 (UTC)

Hurwitz integers ?
I don't see the point of introducing Hurwitz quaternions here (vs. Lipschitz quaternions). If $$q$$ is any Hurwitz quaternion, then $$2 q$$ is Lipschitz and yields the same rational rotation. I suggest removing the Hurwitz reference.

Pascalromon (talk) 09:52, 5 March 2015 (UTC)

Must a,b,c be positive?
As the introduction is currently written, tuples such as $$(-3, 0, 4, 5)$$ would be valid Pythagorean quadruples. I assume that is not intended -- or are these valid but "primitive" P.Q.s? --188.96.85.12 (talk) 08:42, 11 February 2016 (UTC)

One "picture" case is missing:
6²+6²+3²=9²Becker-Sievert (talk) 08:18, 4 March 2016 (UTC)
 * This is not a primitive solution since 3 divides each term. --Bill Cherowitzo (talk) 17:10, 27 April 2017 (UTC)

Parametrization doesn't yield primitives
The article claims, that the first parametrization yields primitive quadruples - but this isn't true. For example:

m=1, n=3, p=1, q=2 → a=5, b=10, c=10, d=15.

m=1, n=5, p=2, q=3 → a=13, b=26, c=26, d=39.

m=1, n=7, p=2, q=1 → a=45, b=30, c=10, d=55.

m=1, n=7, p=3, q=4 → a=25, b=50, c=50, d=75.

m=1, n=7, p=6, q=3 → a=5, b=90, c=30, d=95.

But as far as I can see, the formula seems to correctly generate pythagorean quadruples.

Wayne523 (talk) 08:45, 3 June 2017 (UTC)


 * Indeed. This appears to be an error in our article, not in the references; the Spira reference gives a list of 9 additional conditions one could add to the parameters (m, n, p, q) so that each primitive triple is contained once and only once.  (Spira cites a paper of F. Steiger in the journal Elem. Math. from 1956 for these conditions.)  I'll see what I can do about the wording.  --JBL (talk) 22:40, 3 June 2017 (UTC)