# Can we find $n$ Pythagorean triples with a common leg for any $n$?

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 $n$?

• 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 2:08