<s>[INST] Are there any 2 primitive pythagorean triples who share a common leg? So is it possible for:
$\gcd(a,b,c)=1$
$a^2+b^2=c^2$
and
$\gcd(a,d,e)=1$
$a^2+d^2=e^2$
?
 [/INST] We can find different triples with the same odd legs, if they exist, using this function of $(m,A)$:
$$\text{We can let }n=\sqrt{m^2-A}\text{ where m varies from }\lceil\sqrt{A}\space\rceil\text{ to }\frac{A+1}{2}$$
There is always a Pythagorean triple with side $A=(2m-1),m\in\mathbb{N}\land m\ge2$ and sometimes more than one with the same side length. The smallest of these where $A^2+B^2=C^2\land A^2+D^2=E^2$ is $A=15$ such as in $(15,8,17)$. 
$$m_{min}=\lceil\sqrt{15}\space\rceil=\lceil 3.872983346\rceil=4\text{ and }m_{max}=\frac{A+1}{2}=\frac{16}{2}=8$$
Testing for $m=4\text{ to }8$, we find integers when $(m,n)=(4,1)\text{ and }(8,7)$
$$\text{For }(4,1)\quad A=4^2-1^1=15\quad B=2*4*1=8\quad C=4^2+1^2=17$$
$$\text{For }(8,7)\quad A=8^2-7^2=15\quad D=2*8*7=112\quad E=8^2+7^2=113$$
Sometimes there is only one match such as the smallest $(3,4,5)$ where $m_{min}=m_{max}=2.$ At other times, there are many matches such as for $A=105$ where $m_{min}=11$ and $m_{max}=53.$ and we find $(11,4), (13,8), (19,16),\text{ and }(53,52).$ For these respective values of $(m,n)$ we have:
$$A,B,C=105,88,137\quad D,E=208,233\quad F,G=608,617\quad H,I=5512,5513$$
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