Three primitive Pythagorean triples with the same c Can someone give me an example of Three primitive Pythagorean triples with the same c?
 A: This is a list of the first Pythagorean triples. $1105$ actually has four primitive breakdowns as $a^2+b^2$, and my skimming did not find any earlier ones with three.
$$\begin{align}
1105^2
&=47^2+1104^2\\
&=264^2+1073^2\\
&=576^2+943^2\\
&=744^2+817^2
\end{align}$$
(Maybe @AndreNicolas's answer can be explored more to prove that the number of breakdowns for any $c$ that is the hypotenuse of a Pythagorean triple is a power of $2$.)
A: We produce an answer with hypotenuse $c=5\cdot 13\cdot 17$. So we need to find three relatively prime pairs $\{m,n\}$ of opposite parity such that $m^2+n^2=5\cdot 13\cdot 17$. 
It is simplest to calculate using complex numbers, although the same goal can be achieved by using the Brahmagupta Identity.  Note that $c$ factors over the Gaussian integers as $(2+i)(2-i)(3+2i)(3-2i)(4+i)(4-i)$. 
First one: Note that $(2+i)(3+2i)(4+i)=9+32i$. That gives $m=9$, $n=32$. 
Second one: Do the same calculation using $(2-i)(3+2i)(4+i)$.
Third one: Same calculation, using $(2+i)(3-2i)(4+i)$. 
Remark: There is a fourth, using $(2+i)(3+i)(4-i)$. For an example with $8$ instead of $4$, we can play the same game with $5\cdot 13\cdot 17\cdot 29$. 
A: André Nicolas found the solutions of $a^2+b^2=1105$.  Question was solutions for $c = 1105$ not $c$ being square root of $1105$.  To get the solution of the equations you will need to square each of André Nicolas' complex answers.  For example, his $9+32i$ solution, if you square it you get $-943+576i$, which was the 3rd solution of alex.jordan's solution.  The other $3$ are related in the same method.
A: The answer by alex jordan gave an example of four triples where $C=1105$ and all are primitives.  More can be generated from the divisors of $1105$ and then multiplied by the respective co-factors. E.g. there are $9$ additional non-primitive triples where $C=1105$.
$$221\times(3,4,5)\quad 85\times (5,12,13)\quad 65\times(15,8,17)\quad 17\times(33,56,65)\quad 17\times(63,16,65)\quad 13\times(13,84,85)\quad 13\times(77,36,85)\quad 5\times(21,220,221)\quad 5\times(171,140,221)$$
To find multiple triples for $some$ $C$ we can try different values of $C=4t+1, t\in\mathbb{N}$ as shown below.
Given Euclid's formula $A=m^2-n^2, B=2mn, C=m^2+n^2$
$C= m^2+n^2\implies\space n=\sqrt{C-m^2}$ where $\biggl\lceil\sqrt{\frac{C}{2}}\space\space\biggr\rceil \le m\le\bigl\lfloor\sqrt{C}\bigr\rfloor\quad m,n\in\mathbb{N}$
For example, if $C=65,\biggl\lceil\sqrt{\frac{65}{2}}\space\space\biggr\rceil=6 \le m\le\bigl\lfloor\sqrt{65}\bigr\rfloor=8$ and we find
$$\sqrt{65-6^2}\notin \mathbb{N}\qquad\qquad \sqrt{65-7^2}=4\qquad\qquad \sqrt{65-8^2}=1$$
Our two triples are:
$$f(7,4)=(33,56,65)\quad and \quad f(8,1)=(63,16,65)$$
The larger the value of $C$, the more likely it is to be part of multiple primitives but the next one up from the $1105$ sample has only $1.\quad$ Note: $(1105=4\times276+1)\quad (1109=4\times277+1)$ 
$$(141,1100,1109)$$
