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

Using Wolfram Alpha to ensure accurate algegra, we start with Euclid's formula (modified) for the OP.
$$A=m^2-n^2\qquad B=2mn+(m^2-n^2)\qquad C=m^2+n^2$$
$$A^2=m^4 - 2 m^2 n^2 + n^4$$
$$B^2=m^4 + 4 m^3 n + 2 m^2 n^2 - 4 m n^3 + n^4$$
$$C^2=m^4 + 2 m^2 n^2 + n^4$$
$$A^2+B^2-C^2=m^4 + 4 m^3 n - 2 m^2 n^2 - 4 m n^3 + n^4=0$$
The only solutions to the last equation are:
$$n =\pm m\qquad n = 2 m \pm \sqrt{5} m\qquad n=m=0$$
$\therefore$ there are no integer solutions except $A,B,C=(0,0,0), (0,2,2), (0,8,8), (0,18,18), ...$
