It is trivial if I translate true. We call length of side oppisite A is a Suppose $ S_1(a_1,b_1,c_1),S_2(a_2,b_2,c_2)$ Where $ c_1,c_2$ is the length of longest length side in each triangle. Apply Pytagore law we has: $ a_1^2+b_1^2=c_1^2$ $ a_2^2+b_2^2=c_2^2$ Imply that if two length is equal then the first so . It mean that there are not two triangle satisfy condition.

I can't understand your solution...Could you please explain it in details

Case1$ (a_1,b_1)=(a_2,b_2)$ then $ c_1=c_2$ Case 2$ c_1=c_2$ and one of $ (a_1,b_1)$ equal $ (a_2,b_2)$ Suppose $ a_1=a_2$ then $ b_1=b_2$ It mean that if two length is equal then the third so.

That's wrong, you can e.g. still have $ a_1=c_2$ and $ b_1=b_2$.

ZetaX wrote: That's wrong, you can e.g. still have $ a_1 = c_2$ and $ b_1 = b_2$. I am very sorry. Now consider the case : $ a_1=c_2,b_1=b_2$ (the other case same method and with this case we don't lose the general) $ a_1^2+b_1^2=c_1^2$ $ a_1^2-b_1^2=a_2^2$ So $ a_1^4-b_1^4=(a_2c_1)^2$ This equation has no solution in positive integer (Fermat theorem).