Is there any pattern to pythagorean triples where there are two a-b pairs for one c? I've found lots of Pythagorean triples like this, where there are two triples with the same c. Is there any pattern to them
$17^2 + 144^2 = 145^2$
$24^2 + 143^2 = 145^2$
examples: (by c)
145, 185, 205, 221, 265, 305, 365, 377, 425, 445, 481, 485, 493, 505, 533, 545, 565, 629, 685, 689, 697... the list goes on.
A lot, but not all seem to be multiples of five, and most that are not have two repeated digits, but there are exceptions, like 629, 689, and 697... 
 A: The Brahmagupta–Fibonacci identity says that every product of two sums of two squares is a sum of two squares in two different ways.  So you have
$$
3^2+4^2=5^2\text{ and }20^2+21^2=29^2
$$
so consequently $5^2\times 29^2=145^2$ must be a sum of two squares in two different ways.
A: Suppose $a_1^2 + b_1^2 = c_1^2$ and $a_2^2 + b_2^2 = c_2^2$ are any two solutions. Multiply the first equation by $c_2^2$ and the second equation by $c_1^2$ to get $(a_1c_2)^2 + (b_1c_2)^2 = (a_2c_1)^2 + (b_2c_1)^2 = (c_1c_2)^2$.
If you want only Pythagorean triples that are relatively prime, then consider $(d, e)$ where $d$ and $e$ are relatively prime and of opposite parity. If you find two such pairs such that $d_1^2 + e_1^2 = d_2^2 + e_2^2$ (the common value need not be a square), then $(2e_id_i, e_i^2-d_i^2, e_i^2 + d_i)^2$ for $i=1,2$ are relatively-prime Pythagorean triples with a common value for $c$.
Your first example comes $8^2 + 9^2 = 145 = 1^2 + 12^2$. 
Once you've found one such pair of Pythagorean triples, you can use their $(a_i,b_i)$ as $(d_i,e_i)$ to find another. For example, from $17^2+144^2 = 24^2 + 143^2$, we get $(2 \cdot 17 \cdot 144, 144^2-17^2, 144^2 + 17^2)$ and $(2 \cdot 24 \cdot 143, 143^2 - 24^2, 143^2+24^2)$ are also relatively-prime Pythagorean triples with a common sum.
A: If an odd number $c$ is the hypothenuse of two different primitive Pythagorean triplets, then $c$ is composite, and its prime factors are of the form $4k+1$. The smallest vakue for $c$ is $65=5\cdot 13$. Note that $65=8^2+1^2=7^2+4^2$.
The proof of this fact that I know involves Gaussian integers.
A: There are $\,2^{n-1}\,$ primitive Pythagorean triples for any valid $\,C$-value (hypotenuse) where $\,n\,$ is the number of distinct prime factors of $\,C.\quad$ There may be additional "imprimitives" but the primitives are the interesting ones. This pattern can be seen here where hypotenuse values are repeated when they are composed of multiple primes.
