Pythagorean triples (uniqueness) Is there an elementary way of proving that it is impossible to have $2$ distinct Pythagorean triples $(a_1,b_1,c)$ and $(a_2,b_2,c)$ where $c$ is the same hypothenuse for both triples? Any geometric proof? 
 A: bad news.
$16, 63, 65$
$33, 56, 65$
There are also elementary pairs for $85$, $145$, $185$, $205$, $265$, $305$, $325$, $365$, $377$ (just in case you were wondering if the pattern holds), $425$, $445$, $481$, $485$, and $493$, and that's just the ones below $500$.  Pairs including non-elementary triples start at $c=25$: $7, 24, 25$ and $15, 20, 25$.
A: To make Pythagorean triplets with common hypotenuse, do the following steps:


*

*Suppose you want 3 triplets having common hypotenuse; 

*Take 3 Pythagorean triplets whose sides are coprime to each other.
For eg: (3,4,5), (12,5,13) and (24,7,25)
-Take the Least common multiple of the hypotenuse.
Here it is 325.
-Take the three ratios of the triplets:
(3:4:5), (12:5:13), (24:7:25)
-Take these ratios and make triplets taking 325 as the hypotenuse. They will be Pythagorean triplets.
(195,260,325), (300,125,325),
    (312,91,325)
A: Your hypothesis cannot be proven but it can be disproven by example.
We can find dissimilar triples with matching hypotenuse sides, if they exist, using the function below to find the right $(m,n)$ combinations for Euclid's formula:
$$C=m^2+n^2\Rightarrow n^2=C-m^2\Rightarrow n=\sqrt{C-m^2}$$
$$\text{The function }n=\sqrt{C-m^2}\text{ yields an imaginary number if } m^2>C\text{ so }m_{max}=\sqrt{m}.$$
Whenever we get integer for $n$ and $n<m$ we have the $(m,n)$ needed to find a Pythagorean triple with a hypotenuse equal to $C$. The search is limited to where $\mathbf {1\le m\le\lfloor\sqrt{C}\rfloor}$. For example, we want to find one or more triples with $C=697$. Then $m_{max}=\lfloor\sqrt{697}\rfloor=26.$
Trying different values where $m=1\text{ to }26$, we find $(m,n) = (21,16)\text{ and }(24,11).$
$$A=m^2-n^2\qquad\qquad  B=2mn\qquad\qquad  C=m^2+n^2$$
$$21^2-16^2=185\qquad2*21*16=672\qquad21^2+16^2=697$$
$$24^2-11^2=455\qquad2*24*11=528\qquad24^2+11^2=697$$
Sometimes there are no triples that match, such as if we were to find no integers in our search from $1\text{ to } 26.$ At other times there will be only one match or many such as in the example below.
$$(47,1104,1105)\quad(817,744,1105)\quad(943,576,1105)\quad(1073,264,1105)$$
Of course the surest way of finding matches is to divide each $C$ into $LCM(C_1,C_2,C_3,...)$ and multiply each respective triple by the cofactor for that hypotenuse.
