Generalised Pythagorean Triples

A generalised Pythagorean triple of the form
$$a^2 + b^2 = c^2$$
where $a, b, c \in \mathbb{R}^+$ and $c \geq 2a$, is said to be “normalized” if $b = a$. That is
$$a^2 + b^2 = c^2$$
$$a^2 + b^2 = c^2$$
$$a^2 + a^2 = c^2$$
$$a^2 = c^2$$
$$b = a$$
I have been looking at generalised Pythagorean triples, and they have a relation to regular Pythagorean triples. Specifically, if we consider a generalised Pythagorean triple
$$a^2 + b^2 = c^2$$
Then there exists a primitive Pythagorean triple with the same values except that $2a$ is a factor instead of $a$. However, I’m not sure how to show that if a generalised triple is normalized, then it is primitive.
Any insight into this would be welcome.

A:

Maybe an example will help. Suppose we have a generalised Pythagorean triple $(a^2 + b^2, a^2 + b^2, c^2)$ with $a > b > c > 0$. Then by appropriate rescaling of $a$ and $b$, we may assume that $a = b = 1$. Then
$$1^2 + 1^2 = 2^2 = c^2,$$
$$2^2 + 1^2 = 3^2 = 6,$$
and
3^2 + 1^2 = 4 https://greenearthcannaceuticals.com/wp-content/uploads/2022/06/IconXpert.pdf

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