Can two pythagoras triplet have a common number If I have a pythagoras triplet $(a,b,c)$ such that 
$$a^2+b^2=c^2$$
then is there another triplet $(a,d,e)$ possible such that
$$a^2+d^2=e^2, \; b\neq d$$
 A: Let's rewrite the equation as
$$a^2=(c+b)(c-b)$$
Different factorizations of $a^2$ lead to different triples.  For example, if $a=15$, try $c+b=225$ and $c+b=25$. 
A: You can easily find nonprimitive triples that share an element: start with your two favorite triples (say $3^2 + 4^2 = 5^2$ and $5^2 + 12^2 = 13^2$). We'll get triples with $15$ in each: multiply the first triple by $5^2$ and the second by $3^2$. Then we have $5^2(3^2 + 4^2) = 5^2\cdot5^2$ and $3^2(5^2 + 12^2) = 3^2\cdot13^2$. Hence, $15^2 + 20^2 = 25^2$ and $15^2 + 36^2 = 39^2$. However, you can also find primitive triples sharing an element, for example $20^2 + 21^2 = 29^2$ and $20^2 + 99^2 = 101^2$.
Given an arbitrary triple $a^2 + b^2 = c^2$, there may or may not be another triple $a^2 + d^2 = e^2$ with $e\neq c$. To see this, note that all pythagorean triples are of the form $a = n^2 - m^2$, $b = 2mn$, $c = n^2 + m^2$, where $n,m\in\mathbb{N}$. There is only one pair $(n,m)$ such that any of these three numbers is $3$: $n = 2$, $m = 1$ (this is simple to see if you look at the way differences of squares behave and eliminate the possibility that $n^2 + m^2 = 3$).
A: Each of $\quad A,B,C\quad$ will have as many primitive triples as its number of unique prime factors.
Triples are often generated with Euclid's formula
$$\text{where} \quad A=m^2-k^2 \quad B=2mk \quad C=m^2+k^2$$
$A=3\times5=15\longrightarrow 
f(4,1)=(15,8,17)\quad 
f(8,7)=(15,112,113)$
$B=2^2\times3=12\longrightarrow 
f(3,2)=(5,12,13)\quad 
f(6,1)=(35,12,37)$
$C=5\times13=65\longrightarrow 
f(7,4)=(33,56,65)\quad 
f(8,1)=(63,16,65)$
There are no tiple pairs where two numbers a shared so there is no such thing as
$\quad a^2+b^2=c^2\quad a^2+d^2=c^2\quad b\ne d$
