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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 Dec 9 '16 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 Dec 9 '16 at 10:46
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    $\begingroup$ $1105^2 = 1104^2 + 47^2 = 1073^2 + 264^2 = 943^2 + 576^2 = 817^2 + 744^2$ $\endgroup$ – achille hui Dec 9 '16 at 11:08
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    $\begingroup$ This is a nice question +1 $\endgroup$ – Vidyanshu Mishra Dec 9 '16 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 Dec 9 '16 at 16:56
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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? $\endgroup$ – Atul Mishra Dec 9 '16 at 11:41
  • $\begingroup$ Ahhh, you are right. $\endgroup$ – achille hui Dec 9 '16 at 11:42
  • $\begingroup$ More examples here $\endgroup$ – Ng Chung Tak Dec 9 '16 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 $\endgroup$ – Atul Mishra Dec 9 '16 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. $\endgroup$ – Ilmari Karonen Dec 9 '16 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. $\endgroup$ – Shraddheya Shendre Dec 9 '16 at 14:51
  • $\begingroup$ My answer now provides a counterexample where the numbers are pairwise coprime. $\endgroup$ – lokodiz Dec 9 '16 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 Dec 9 '16 at 14:40
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    $\begingroup$ @Sophie Nice analysis $\endgroup$ – yoyostein Dec 9 '16 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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    $\begingroup$ $127^2 + 236^2 = 1^2 + 268^2 = \cdots = 17\times 65^2 \ne 1105^2$. $\endgroup$ – achille hui Dec 9 '16 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. $\endgroup$ – DanielWainfleet Dec 9 '16 at 18:27
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Euclid's formula will generate all triples that exist as primitives, doubles or square multiples of primitives. Using this knowledge and the fact that side $B$ is always a multiple of $4$, we can start with a triple where $A,B,C$ are all multiples of $4$ such as $(12,16,20).$ Let $x=20$.

If we solve any of Euclid's formula functions for $n$, we can find triples for any given side, if they exist, with a finite search of $m$ values while looking for which $m$ values yield an integer $n$.

$$\text{For }A=m^2-n^2\text{, we let } n=\sqrt{m^2-A}\text{ : }\lceil\sqrt{A+1}\space\rceil\le m\le \biggl\lceil\frac{A}{2}\biggr\rceil$$.

$$n=\sqrt{m^2-20}\text{ where }\lceil\sqrt{20+1}\space\rceil=5\le m\le \biggl\lceil\frac{20}{2}\biggr\rceil=10$$ We find $m=6$ yields $n=4$ but $m=5,7,8,9,10$ yield non-integers... $f(6,4)=(20,48,52)$

$$\text{For }B=2mn,n=\frac{B}{2m}\text{ where }\lceil\sqrt{B}\space\space\rceil\le m \le \frac{B}{2}$$

$$n=\frac{20}{2m}\text{ where }\lceil\sqrt{20}\space\rceil=5\le m \le \frac{20}{2}=10$$ We find $\qquad f(5,2)=(21,20,29)\qquad f(10,1)=(99,20,101)$

$$\text{For }C= m^2+n^2,\space n=\sqrt{C-m^2}\text{ where }\biggl\lceil\sqrt{\frac{C}{2}}\space\space\biggr\rceil \le m\le\bigl\lfloor\sqrt{C}\bigr\rfloor$$.

$$n=\sqrt{20-m^2}\text{ where }\biggl\lceil\sqrt{\frac{20}{2}}\space\space\biggr\rceil=4 \le m\le\bigl\lfloor\sqrt{20}\bigr\rfloor=4$$ We find $f(4,2)=(12,16,20)$

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