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Just to be clear, I call an integer $c$ a 'primitive Pythagorean hypotenuse' if there exist coprime integers $a$ and $b$ satisfying $a^2+b^2 = c^2$. I noticed that the set of such primitive Pythagorean hypotenuses seems to form a closed set under multiplication (and indeed this is backed up by this Mathworld page -- see equations (25)-(26)). However, I can't find a proof of it (the link above provides a book reference but I don't have access to it), or come up with one myself.

Some thoughts: Euler showed that $c$ is a primitive Pythagorean triple iff it is odd and can be written as the sum of two squares $c = m^2+n^2$ for some coprime integers $m$ and $n$. It is straightforward to show that the product of a sum of two squares is also a sum of two squares (and hence the set of all Pythagorean hypotenuses is closed under multiplication), but I'm having trouble with the primitive/coprime part. In other words I can show that if $c_1 = m_1^2 + n_1^2$ and $c_2 = m_2^2 + n_2^2$ then there exist $m_3$ and $n_3$ such that $c_1c_2 = m_3^2 + n_3^2$, but I can't show that if $gcd(m_1,n_1)=gcd(m_2,n_2)=1$ then it is possible to choose $m_3$ and $n_3$ such that $gcd(m_3,n_3)=1$. Any hints or references would be appreciated!

Edit (in response to individ's comments): Note that the formulae $$ m_3 = m_1 n_2 - n_1 m_2 \ , \ n_3 = m_1 m_2 + n_1 n_2 $$ and $$ m_3 = m_1 n_2 + n_1 m_2 \ , \ n_3 = m_1 m_2 - n_1 n_2 $$ do not always produce coprime $m_3$ and $n_3$. The earliest example of this I could find is $$ m_1 = 8, n_1 = 1, m_2 = 31, n_2 =92 $$ for which the first formula above gives $m_3 = 705, n_3 = 340$ (both divisible by 5), and the second formula above gives $m_3=767, n_3=156$ (both divisible by 13). However $m_3^2 + n_3^2 = 612,625$ is still the hypotenuse of a primitive triple because $$ 199176^2 + 579343^2 = 612625^2 $$ and $gcd(199176,579343)=1$. (Sorry for the rather ugly example, but it was the simplest I could find (maybe because I don't know how to search efficiently!).)

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  • $\begingroup$ what do you mean by multiplication? are you saying that if you define $(a,b,c).(d,e,f) = (ad,be,cf)$ then the product of two primitive triples is another one? I really doubt that this is true. $\endgroup$ – Mark Joshi Jan 6 '15 at 8:39
  • $\begingroup$ This is possible if at least one hypotenuse squared. $$c_1^2=m_1^2+n_1^2$$ Then you can make $$c_1^2c_2=(c_1m_2)^2+(c_1n_2)^2$$ On the other need to write a system of equations and try to solve it. Write what system is better? $\endgroup$ – individ Jan 6 '15 at 9:44
  • $\begingroup$ I think to solve the problem in General we have to solve the Diophantine equation. $$(m^2+n^2)(x^2+y^2)=z^2+q^2$$ The solution there. mathoverflow.net/questions/181242/… The formula is there and written. $\endgroup$ – individ Jan 6 '15 at 10:04
  • $\begingroup$ @MarkJoshi: I mean multiplying the hypotenuses only: i.e. show that $cf$ is the hypotenuse of another primitive Pythagorean triple (the 'legs' don't have to be $ad$ and $be$). $\endgroup$ – Mark A Jan 6 '15 at 12:47
  • $\begingroup$ @individ: Regarding your first comment, that solution is no use because the resulting Pythagorean triple is not primitive ($c_1^2$ is a common factor). Regarding your second comment, it's not enough to solve the Diophantine equation, you also have to show that $z$ and $q$ can be chosen to be coprime (which is the bit I'm struggling with). $\endgroup$ – Mark A Jan 6 '15 at 12:49
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Let consider the ring of Gaussian integers $\mathbb Z[i]$. For $\zeta=a+ib\in \mathbb Z[i]$ we say that $\zeta$ is primitive if $\gcd(a,b)=1$. Note that $\zeta$ is primitive if and only if for each $d\in\mathbb Z$, $d\mid\zeta$ implies $d=\pm 1$. Consequently, factors of primitive numbers are primitive.

An odd integer $x\in\mathbb Z$ is a primitive Pythagorean hypotenuse if and only if there exists a primitive $\xi\in\mathbb Z[i]$ such that $x=\xi\bar\xi$.

Let $x=\xi\bar\xi$ be a primitive Pythagorean hypotenuse and, since $\mathbb Z[i]$ is an UFD, take the prime factorization $\xi=\pi_1^{e_1}\cdots \pi_r^{e_r}$. Since $\xi$ is primitive and $x$ odd, each factor $\pi_j$ is primitive and with no common factor with $2$, hence $\Im(\pi_j^4)\neq 0$. Since $\pi_j\bar\pi_j\mid x$ can assume that $\Im(\pi_j^4)>0$ for each $j$.

Let $y=\eta\bar\eta$ be another primitive Pythagorean hypotenuse with $\eta=o_1^{d_1}\cdots o_s^{d_S}$ and $\Im(o_j^4)>0$.

We claim that $\xi\eta$ is primitive. For assume on contrary that $d\mid\xi\eta$ for some $d\in\mathbb Z$. Let $\pi\in\mathbb Z[i]$ be an irreducible factor of $d$. Then, wlog, $\pi\mid\xi$. Then $\pi$ is primitive and $\bar\pi\mid d\mid\xi\eta$, but $\bar\pi\not\mid\xi$, hence $\bar\pi\mid \eta$. This is a contradiction, because $\Im(\bar\pi^4)<0$.

This proves that $xy$ is a primitive Pythagorean hypotenuse.

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  • $\begingroup$ This looks good - thanks! Just one question: I think the third sentence should be "$\zeta$ is not primitive iff it's divisible by an integer in $\mathbb{Z}$" (rather than $\mathbb{Z}[i]$). Is this right? $\endgroup$ – Mark A Jan 7 '15 at 4:43
  • $\begingroup$ Yes, I mean: $\zeta$ is not primitive iff it's divisible in $\mathbb Z[i]$ by an integer in $\mathbb Z$. $\endgroup$ – Fabio Lucchini Jan 7 '15 at 10:41
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Instead of multiplying just the hypotenuses multiply two primitive triples using complex numbers to get a third one.

$\begin{array}{l} ({a_1},{b_1}i,{c_1})({a_2},{b_2}i,{c_2}) = ({a_3},b{i_3},{c_3})\\ {a_3} = ({a_1}{a_2} + {b_1}{b_2}{i^2})\\ {b_3} = \left( {{a_1}{b_2}i + {a_2}{b_1}i} \right)\\ {c_3} = \left( {{c_1}{c_2}} \right) \end{array}$

There are no common factors so $({a_3},{b_3},{c_3})$ is primitive.

$\begin{array}{l} (3,4i,5)(5,12i,13) = ({a_3},b{i_3},{c_3})\\ {a_3} = (15 - 48) = - 33\\ {b_3} = \left( {36i + 20i} \right) = 56i\\ {c_3} = {65} \end{array}$

$\left( { - {{33}^2} + 56{i^2} = {{65}^2}} \right) \to {33^2} + {56^2} = {65^2}.$

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  • $\begingroup$ The statement that $a_3$ and $b_3$ share no common factors is false in general -- see the example I give at the end of the question. Hence this method does not always produce a new primitive triple. $\endgroup$ – Mark A Feb 20 '15 at 9:25
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Rewrite the equation in this form:

$$(m^2+n^2)(x^2+y^2)=z^2+q^2$$

Then the solution is always there.

$$z=mx+ny$$

$$q=nx-my$$

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  • $\begingroup$ But you have to show why $z,q$ are relatively prime. $\endgroup$ – Wojowu Jan 6 '15 at 10:59
  • $\begingroup$ @Wojowu It's not all decisions. I wrote the most simple. To obtain relatively simple solutions have to choose a certain way factors. It's simple arithmetic - is not interesting to me. $\endgroup$ – individ Jan 6 '15 at 11:04
  • $\begingroup$ @individ: thanks for your answer, but as I commented above, showing that $z$ and $q$ are coprime is what I find difficult (and is not always true for the specific $z$ and $q$ you give here). $\endgroup$ – Mark A Jan 6 '15 at 12:50
  • $\begingroup$ @MarkA You substitute $(m,n) - (x,y)$ different parity and see what happens. $$(m,n) - (5,2)$$ $$(x,y) - (7,4)$$ $$(z,q) - (43,6)$$ there is no problem. This gives Pythagorean triples. $$20^2+21^2=29^2$$ $$33^2+56^2=65^2$$ $$1813^2+516^2=1885^2$$ $\endgroup$ – individ Jan 6 '15 at 13:18
  • $\begingroup$ @individ I agree your formula always gives Pythagorean triples, but it doesn't always give primitive Pythagorean triples -- please see the counterexample I have added to the question... $\endgroup$ – Mark A Jan 6 '15 at 22:20

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