If $(a,b,M)$ is a Pythagorean triple, can $(b,b+a,N)$ be another triple?

Question

Does anyone know of a pair of Pythagorean triples of the form $$(a, b, M) \quad\text{and}\quad(b, b+a, N)$$ Is such a pair possible?

Answer 1

Partial solution follows:

First note that if $a$ and $b$ share a common factor, then you can divide both triples by that common factor and the property still holds (e.g. if $a$ and $b$ are both even, then so is $a+b$, and hence the triplets $(a/2, b/2, M/2)$ and $(b/2, \frac{a+b}{2}, N/2)$ should also be Pythaogorean). Hence, without loss of generality, assume that $\gcd(a, b) = 1$, and hence $(a, b, M)$ is a primitive triple. Then $(b, a+b, N)$ is also a primitive triple.

Then assume that the generators of $(a, b, M)$ and $(b, a+b, N)$ are $(x_1, y_1)$ and $(x_2, y_2)$ respectively, i.e. that $a$ and $b$ equal $2x_1y_1$ and $x_1^2-y_1^2$ in some order, and similarly for $b$ and $a+b$. That gives four possible cases to test against, but three of them can very quickly be eliminated:

$a+b$ is always odd (because exactly one of $a$ and $b$ is even), so it must be that $b = 2x_2y_2$ and $a+b = x_2^2 - y_2^2$. Hence $b$ is even, so $a$ is odd, hence $a = x_1^2 - y_1^2$ and $b = 2x_1y_1$, implying that $x_1y_1 = x_2y_2$.

We can then rearrange all of that to eliminate $a$, $b$ and $y_2$, and hence have an expression relating $x_1$, $y_1$ and $x_2$:

$x_2^4 - (x_1^2 - y_1^2 + 2x_1y_1)x_2^2 - x_1^2y_1^2 = 0$

And we note that in this form, it is a quadratic in $x_2^2$, so the triples exist only if that quadratic has at least one solution that is a perfect square. An equivalent equation can be written for $y_2$. So for any given Pythagorean triple, there is a means of testing if it will form a second triple, but I don't know of any way to simplify that or prove that it will never have a solution.