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John Leech has a nice paper entitled, "Two Diophantine birds with one stone". The two birds in question are the two systems, $$t^2−3\big(a^2, b^2, (a + b)^2, (a−b)^2\big) = p^2, q^2, r^2, s^2$$ $$u^2 + (c^2, d^2, (c + d)^2, (c − d)^2) = p^2, q^2, u^2, v^2$$

So Leech was able to solve simultaneously the problem of $n=4$ Pythagorean triples with a common leg $a$,

$$a^2+b_1^2 = c_1^2,\quad a^2+b_2^2 = c_2^2\\a^2+b_3^2 = c_3^2,\quad a^2+b_4^2 = c_4^2$$

using an elliptic curve, hence there were infinitely many solutions.

The most I've known is $n=10$ triples found by R. Rathbun with the common leg,

$$a =232792560 = 2^4·3^2·5·7·11·13·17·19$$

and $10$ $b_n$ as, $$\color{blue}{55306628},\; \color{green}{117515475},\; 71608131,\; 135412420,\; 135423925,\; 447886692,\; 153939420,\; 414785371,\; 180609955,\; 1219785588$$ so,

$$a^2 + \color{blue}{55306628}^2 = 239272228^2\\ a^2 + \color{green}{117515475}^2 =260772435^2\\ \vdots$$ and so on. (P.S. This $a$ is not necessarily the smallest for $n=10$.)

Q: Can we in fact solve a system of $n$ Pythagorean triples with a common leg $a$ and $n$ distinct $b_n$ for any $n$?

Tito Piezas III
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    Due to this reddit thread (of all places!) that I just found, I figured out how to answer my question. But I'll refrain from answering since I'm sure the solution will also occur to someone else. – Tito Piezas III Jan 13 '18 at 14:36
  • Easy happy task. It all comes down to the decomposition of numbers into factors. After decomposition, find the desired number and building Pythagorean triples. $$a=2ps=2kt=(x-y)(x+y)=(z-v)(z+v)=.......$$ – individ Jan 13 '18 at 15:05
  • More easier: $;{{\left( {{a}^{2}}, {{b}^{2}}-{{c}^{2}}\right) }^{2}}+4 {{a}^{2}}, {{b}^{2}}, {{c}^{2}}$,$;{{\left( {{a}^{2}}-{{b}^{2}}, {{c}^{2}}\right) }^{2}}+4 {{a}^{2}}, {{b}^{2}}, {{c}^{2}}$,$;{{\left( 1-{{a}^{2}}, {{b}^{2}}, {{c}^{2}}\right) }^{2}}+4 {{a}^{2}}, {{b}^{2}}, {{c}^{2}}$,$;{{\left( {{a}^{2}}, {{c}^{2}}-{{b}^{2}}\right) }^{2}}+4 {{a}^{2}}, {{b}^{2}}, {{c}^{2}}$. Adding d^2, g^2 etc. get more equalities with a common leg $2abcdg...$ – AlexSam Jan 14 '18 at 02:08
  • @Tito Piezas III There is a matching side A for any odd number $>1$ but not so for side B or C. There is a simple way to find which values of $n$ have a corresponding value in a triple where $GCD(A,B,C)=(2m-1)^2,m\in \mathbb{N}$. See matching sides of Pythagorean triples. – poetasis May 27 '19 at 17:07

1 Answers1

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We can find Triples by matching side-A to any odd number greater than one and, with multiples, any other natural number. There are also one or more triples with side-B for any multiple of four. Let's start with side-A by solving the Euclid's formula $ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$ side-A for k.

\begin{equation} A=m^2-k^2\implies k=\sqrt{m^2-A}\qquad\text{for}\qquad \sqrt{A+1} \le m \le \frac{A+1}{2} \end{equation} The lower limit ensures $k\in\mathbb{N}$ and the upper limit ensures $m> k$.

Now, let $A=3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19=4849845 $

$$A=4849845\implies \sqrt{4849845+1}=2202\le m \le \frac{4849845+1}{2} =2424923 $$ There $128$ factors of $4849845$ and perhaps as many Pythagorean triples where $A=4849845$. Here are the first $34$ generated by a spreadsheet using this formula in Excel.

\begin{equation}\qquad f(2203,58)=(4849845,255548,4856573)\qquad f(2213,218)=(4849845,964868,4944893)\quad f(2251,466)=(4849845,2097932,5284157)\quad f(2267,538)=(4849845,2439292,5428733)\quad f(2309,694)=(4849845,3204892,5813117)\quad f(2341,794)=(4849845,3717508,6110717)\quad f(2389,926)=(4849845,4424428,6564797)\quad f(2389,926)=(4849845,4424428,6564797)\quad f(2389,926)=(4849845,4424428,6564797)\quad f(2581,1346)=(4849845,6948052,8473277)\quad f(2677,1522)=(4849845,8148788,9482813)\quad f(2747,1642)=(4849845,9021148,10242173)\quad f(2843,1798)=(4849845,10223428,11315453)\quad f(2923,1922)=(4849845,11236012,12238013)\quad f(2987,2018)=(4849845,12055532,12994493)\quad f(3061,2126)=(4849845,13015372,13889597)\quad f(3643,2902)=(4849845,21143972,21693053)\quad f(3749,3034)=(4849845,22748932,23260157)\quad f(3979,3314)=(4849845,26372812,26815037)\quad f(3989,3326)=(4849845,26534828,26974397)\quad f(4181,3554)=(4849845,29718548,30111677)\quad f(4373,3778)=(4849845,33042388,33396413)\quad f(4603,4042)=(4849845,37210652,37525373)\quad f(5557,5102)=(4849845,56703628,56910653)\quad f(5867,5438)=(4849845,63809492,63993533)\quad f(6277,5878)=(4849845,73792412,73951613)\quad f(6491,6106)=(4849845,79268092,79416317)\quad f(6971,6614)=(4849845,92212388,92339837)\quad f(7669,7346)=(4849845,112672948,112777277)\quad f(8651,8366)=(4849845,144748532,144829757)\quad f(9019,8746)=(4849845,157760348,157834877)\quad f(9637,9382)=(4849845,180828668,180893693)\quad f(9941,9694)=(4849845,192736108,192797117)\quad f(10613,10382)=(4849845,220368332,220421693) \end{equation}

If you want to find triples match C or B, here are some formulas.

\begin{equation} B=2mk\implies k=\frac{B}{2m}\qquad\text{for}\qquad \bigg\lfloor \frac{1+\sqrt{2B+1}}{2}\bigg\rfloor \le m \le \frac{B}{2} \end{equation} The lower limit ensures $m>k$ and the upper limit ensures $m\ge 2$

\begin{equation} C=m^2+k^2\implies k=\sqrt{C-m^2}\qquad\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor \end{equation} The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$.

poetasis
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