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While playing with numbers, I thought about squares of numbers, and then the first thing that came to mind was Pythagorean triplets.

I observed a very interesting fact that any $x\in\mathbb N$ can never be a member of more than two Pythagorean triplets of pairwise coprime numbers, like $(3,4,5)$ and $(8,15,17)$.

For example $$16^2+63^2=65^2$$$$33^2+56^2=65^2$$

are the possible triplets for $x=65$, and $65$ cannot exist in any other triplet of co-primes.

Now I need to prove this conjecture.

So I thought that in a Pythagorean triplet, the three numbers are of the form $(2mn, m^2-n^2, m^2+n^2)$.

Let $x$ be a number . Then I have to show that $$x=2mn$$$$x=a^2+b^2$$$$x=y^2-z^2$$ are not simultaneously possible.

But I am stuck and don't know where to go from here.

Please help me prove this or help me disprove it by giving a counter example.

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    $\begingroup$ The pythagorean triples are uniquely generated by $2kmn, k(m^2-n^2),k( m^2+n^2)$ where $\gcd(m,n)=1, m \gt n\ge 1,k\ge1$ $\endgroup$
    – miracle173
    Commented Dec 9, 2016 at 10:41
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    $\begingroup$ maybe you should post another question with the requirement $(x,y)=(y,z)=(z,x)=1$ $\endgroup$
    – miracle173
    Commented Dec 9, 2016 at 10:46
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    $\begingroup$ $1105^2 = 1104^2 + 47^2 = 1073^2 + 264^2 = 943^2 + 576^2 = 817^2 + 744^2$ $\endgroup$ Commented Dec 9, 2016 at 11:08
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    $\begingroup$ This is a nice question +1 $\endgroup$ Commented Dec 9, 2016 at 11:57
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    $\begingroup$ "solve it again" is not a fair approach. As I already suggested you should open another question if your question is already answered but you you want to change it substantially $\endgroup$
    – miracle173
    Commented Dec 9, 2016 at 16:56

6 Answers 6

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Start from $1105 = 5\cdot 13\cdot 17$ whose prime factors all has the form $4k+1$.
Decompose the prime factors in $\mathbb{Z}$ over gaussian integers $\mathbb{Z}[i]$ as

$$5 = (2+i)(2-i),\quad 13 = (3+2i)(3-2i)\quad\text{ and }\quad17 = (4+i)(4-i)$$

Recombine the factors of $1105^2$ over $\mathbb{Z}[i]$ in different order and then turn them to sum of two squares. We get:

$$\begin{array}{rc:rr} 1105^2 = & 943^2 + 576^2 & ((2+i)(3+2i)(4+i))^2 = & -943 + 576i\\ = & 817^2 + 744^2 & ((2-i)(3+2i)(4+i))^2 = & 817 + 744i\\ = & 1073^2 + 264^2 & ((2+i)(3-2i)(4+i))^2 = & 1073 + 264i\\ = & 1104^2 + 47^2 & ((2-i)(3-2i)(4+i))^2 = & -47 - 1104i\\ \end{array} $$ A counter-example for the speculation that an integer can appear in at most two primitive Pythagorean triples.

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  • $\begingroup$ shouldn't it be 13 in the place of 11@achille hui? $\endgroup$ Commented Dec 9, 2016 at 11:41
  • $\begingroup$ Ahhh, you are right. $\endgroup$ Commented Dec 9, 2016 at 11:42
  • $\begingroup$ More examples here $\endgroup$ Commented Dec 9, 2016 at 23:18
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The "smallest" counterexample is \begin{gather*} 5^2 + 12^2 = 13^2 \\ 9^2 + 12^2 = 15^2 \\ 12^2 + 16^2 = 20^2 \end{gather*} ($12$ also satisfies $12^2 + 35^2 = 37^2$.)

Edit: now that the OP has stipulated that the elements be pairwise coprime, the smallest counterexample (in the sense that the repeated number is minimal) is

\begin{gather*} 11^2 + 60^2 = 61^2 \\ 60^2 + 91^2 = 109^2 \\ 60^2 + 221^2 = 229^2 \end{gather*}

$60$ also satisfies $60^2 + 899^2 = 901^2$.

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  • $\begingroup$ I have added something to the question , please solve it again @lokodiz $\endgroup$ Commented Dec 9, 2016 at 11:00
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$120,160,200$ and $90,120,150$ and $72,96,120$

How I arrived at it: It is a common knowledge that if we scale the triplet $3,4,5$ by any constant, we get another triplet. So I found out a common multiple of $3,4,5$, which is $120$ and then scaled the triplet one by one with the constants $\frac{120}{3}$, $\frac{120}{4}$ and $\frac{120}{5}$.

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  • $\begingroup$ Note that the question has now been edited to allow only triplets whose numbers don't share a common factor. By construction, your examples don't pass that added requirement. $\endgroup$ Commented Dec 9, 2016 at 14:39
  • $\begingroup$ Yes, I saw that. but unfortunately, I am unable to prove/disprove it. But I have a hunch that a counter example can be found. I should maybe start thinking about a proof rather than search for counter example. $\endgroup$ Commented Dec 9, 2016 at 14:51
  • $\begingroup$ My answer now provides a counterexample where the numbers are pairwise coprime. $\endgroup$
    – lokodiz
    Commented Dec 9, 2016 at 15:04
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$16, 63, 65$

$25, 60, 65$

$33, 56 , 65$

Observed from: http://www.tsm-resources.com/alists/trip.html

Possible interesting reasearch question: Are there numbers that appears in infinitely many Pythagorean triples?

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    $\begingroup$ @yoyostein, suppose such a number exists. Call it $n$. Then at least one of $a^2+b^2=n$, $2ab=n$, $a^2-b^2=n$ has infinitely many solutions. The first two obviously don't, and the third has finitely many because $a+b$ and $a-b$ have to be factors of $n$ (a finite set). $\endgroup$
    – Sophie
    Commented Dec 9, 2016 at 14:40
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    $\begingroup$ @Sophie Nice analysis $\endgroup$
    – yoyostein
    Commented Dec 9, 2016 at 14:43
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Any prime which is congruent to $1$ modulo $4$ is the largest member of a Pythagoren triplet. (That is not obvious.)

If $P_1,..., P_n$ are $n$ distinct primes, each congruent to $1$ modulo 4, then $\prod_{j=1}^nP_j$ is the largest member of $2^{n-1}$ different primitive Pythagorean triplets. ( It is not obvious that the method, below, always yields that many.)

Use $(aa'+bb')^2+(ab'-ba')^2=(aa'-bb')^2+(ab'+ba')^2=(a^2+b^2)(a'^2+b'^2).$

Example : (For ease of typing let "$.$" denote "$\times$"). From $3^2+4^2=5^2$ and $5^2+12^2=13^2$ we have $$(3.5\pm 4.12)^2+(3.12\mp 4.5)^2=5^2.13^2.$$ That is, $63^2+16^2=33^2+56^2=65^2.$...... Since $17=1^2+4^2$ we have $$(63.1\pm 16.4)^2+(63.4\mp 16.1)^2=17^2.65^2=$$ $$= (33.1\pm 56.4)^2+(33.4\mp 56.1)^2=17^2.65^2.$$ That is, $$127^2+236^2=1^2+268^2=257^2+76^2=191^2+188^2=1105^2.$$

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    $\begingroup$ $127^2 + 236^2 = 1^2 + 268^2 = \cdots = 17\times 65^2 \ne 1105^2$. $\endgroup$ Commented Dec 9, 2016 at 17:46
  • $\begingroup$ @achillehui. Right. I should have proceeded from $17^2=15^2+8^2$, not from $17=4^2+1^2$. I'll fix it tomorrow. $\endgroup$ Commented Dec 9, 2016 at 18:27
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There are $2^{n-1}$ primitive Pythagorean triples for every valid $C$-value where $n$ is the number of "distinct" prime factors of $C$.

For $C=5*13*17*29=32045 ,\quad$ there are $2^{4-1}=8$ primitives

$$f(131,122)=(2277,31964,32045)\\ f(142,109)=(8283,30956,32045)\\ f(157,86)=(17253,27004,32045)\\ f(163,74)=(21093,24124,32045)\\ f(166,67)=(23067,22244,32045)\\ f(173,46)=(27813,15916,32045)\\ f(178,19)=(31323,6764,32045)\\ f(179,2)=(32037,716,32045)\\ $$

All Pythagorean triples take the form $\quad A=4x^2+1\quad B=4x\quad C=4x+1$ $LCM\big((4x^2+1),\space 4x,\space (4x+1)\big)=100$

For $A=B=C=100,\quad$ there are $3$ triples

\begin{align*} F(26,24)\quad &\rightarrow A= 26^2 - 24^2=100 &\rightarrow (100,1248,1252)&\\ F(10,5)\quad & \rightarrow B= 2(10)(5) = 100 &\rightarrow (75,100,125)&\\ F(8,6)\quad & \rightarrow C= 8^2 + 6^2 = 100 &\rightarrow ( 28,96,100)& \end{align*}

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