TITLE: Pythagorean primitive triples
QUESTION [5 upvotes]: If $a,b,c$ are primitive Pythagorean triples such that $c^2=a^2+b^2$, prove that this is unique primitive Pythagorean triple using $c$ as hypotenuse.
I'm not sure if I phrased this correctly. I want to ask, if $c^2=a^2+b^2$ where $a,b,c$ are pairwise coprime, prove that there are no $a_1$ and $b_1$ such that $a_1,b_1,c$ that satisfy $c^2=a_1^2+b_1^2$, and they are also Pythagorean triple.
This is obviously true based on the short list I found on Wiki but I can't seem to prove it... Help?
EDIT Since this obviously seems false, my question now is, if the process described above false only for $c=5k$? Since it seems to be false for $c=65, 145, 185, 205, 265...$ Can one prove this?

REPLY [4 votes]: Hint $\rm\ \ \ c^2 = N(a+b {\it i}) = a^2 + b^2,\ \ \ C^2 = N(A+B{\it i}) = A^2 + B^2\ $ so by multiplicativity of norms
$\rm\quad\Rightarrow\ (cC)^2 = N(a\pm b{\it i})\:N(A + B{\it i}) = N(aA\mp bB + (aB\pm bA){\it i}) = (aA\mp bB)^2 + (aB\pm bA)^2 $
e.g. $\ \ \ 5^2 = 4^2 + 3^2,\ \ 13^2 = 12^2 + 5^2\ \Rightarrow\ 65^2 =\: (48\mp 15)^2 + (20\pm 36)^2 =\: \{63^2+16^2,\: 33^2+56^2\}$