Is a Pythagorean triple uniquely determined by its smallest element?

Question

A triple $(a,b,c)$ is a Pythagorean triple if $a,b$ and $c$ are strictly positive integers satisfying $a^2 + b^2 = c^2$.

Do there exist $a,b,c,b',c'$ for which $(a,b,c)$ and $(a,b',c')$ are both Pythagorean triples, with $a \le b$ and $a\le b'$, but $b\neq b'$?

[Such examples do exist as commenters below have pointed out.]

Answer 1

Here is an infinite class of primitive Pythagorean pairs.

Give integers $a,b$ with $b>1,$

1. $a>(\sqrt 2+1)b,$ and
2. $\gcd(a,b)=1$ and
3. $a,b$ are not both odd.

Then we have triples:

$$(2ab,a^2-b^2,a^2+b^2)\\ (2ab,a^2b^2-1,a^2b^2+1)$$

Condition $1.$ ensures $a^2-b^2>2ab.$ Conditions $2.$ and $3.$ ensure the triples are primitive.

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The smallest pair is $(a,b)=(5,2)$ which gives the triples listed in comments:

$$(20,21,29)\\(20,99,101)$$