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I know some ways to find some Pythagorean triples.
And I understand that if $a^2 + b^2 = c^2$ then $(a-b)^2 + (a+b)^2 = 2c^2$.
I feel like that suggests a way forward, but I cannot find that way.
Is there an algorithm to pick four whole numbers to serve as the legs of two right triangles such that the triangles are not congruent but the hypotenuses are the same whole number?
Let $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ be arbitrary non-proportional pythagotrean triples, and let $d:={\rm gcd}(c_1,c_2)$. Then $${c_2\over d}\left(a_1,b_1,c_1\right),\qquad{c_1\over d}\left(a_2,b_2,c_2\right)$$ are pythagorean triples to noncongruent right triangles with the same hypotenuse $c={c_1c_2\over d}$.
It is possible to find examples of two or more non-congruent right triangles with the same hypotenuse. I don't know of a systematic way to search for them (I used brute force). A few thoughts directed me to a choice of a hypotenuse to research: the hypotenuse of a primitive right triangle with integer sides must be odd; and that odd number may not have an odd number of prime factors congruent to $3\mod(4)$ if it can be represented as $m^2+n^2$. I chose to look at $1105=5\cdot 13\cdot 17$, which yielded the following three primitive Pythagorean triples: $(1105,1104,47),\ (1105,1073,264),\ (1105,943,576)$.
In the $m^2-n^2,\ 2mn,\ m^2+n^2$ representation of such triples, the first corresponds to $m=24,\ n=23$; the second to $m=33,\ n=4$; and the third to $m=32,\ n=9$.
Added by edit: Fermat proved that every prime number of the form $4n+1$ can be expressed as the sum of two squares in only one way. Primes of the form $4n+3$ cannot be expressed as the sum of two squares.
The Brahmagupta–Fibonacci identity says that $$(a^2 + b^2) (c^2 + d^2) = (a c − b d)^2 + (a d + b c)^2 = (a c + b d)^2 + (a d − b c)^2$$
This means that two suitable primes, each of which can be represented uniquely as the sum of two squares, have a product that can be represented in two distinct ways as the sum of two squares. Thus any odd number that is the product of two distinct primes, each of which has the form $4n+1$, can be the hypotenuse of two distinct primitive Pythagorean triples (as exemplified in the comment by David K). The inclusion of more suitable prime factors in the number corresponding to the hypotenuse will increase the number of ways that the resulting product can be represented as the sum of two squares (as exemplified in my earlier given example).
You know Euclide's formula that gives all primitives pythagorean triples starting from two integers $m>n>0$, and this formula say that we can find only one value $c=m^2+n^2$. Non primitive triples can be found multiplying a primitive by an integer $k$ so, for any $k$ we have different triples and we conclude that we can have only one pythagorean triple for any $c$.
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:
$$n=\sqrt{C-m^2}$$ Whenever we get integer for $n$ and $n<m$ then we have the $(m,n)$ needed to find a Pythagorean triple with a hypotenuse equal to $C$. The search is limited to where $\mathbf {m=1\text{ to }\lfloor\sqrt{C}\rfloor}$. 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 $(21,16)\text{ and }(24,11).$
$$A=m^2-n^2\qquad B=2mn\qquad C=m^2+n^2$$ $$21^2-16^2=185\qquad2*21*16=672\qquad21^2+16^2=697$$ $$24^2-11^2=455\qquad2*24*11=528\qquad24^2+11^2=697$$ Sometimes there are no triples that match, such as if we 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.
$$(47,1104,1105)\quad(817,744,1105)\quad(943,576,1105)\quad(1073,264,1105)$$
