A Pythagorean Triple ?

$a = \dfrac{m}{n}(b + c)$ and $a + b + c = 2n(n + m) \implies b + c = 2n(n + m) - a \implies$

$na = m(2n(n + m) - a) \implies (n + m)a = 2nm(n + m) \implies \boxed{a = 2nm}$

$\implies b + c = 2n(n + m) - 2nm = 2n^2 \implies c = 2n^2 - b \implies$

$(2n^2 - b)^2 = 4n^2m^2 + b^2 \implies 4n^4 - 4n^2b + b^2 = 4n^2m^2 + b^2 \implies$

$n^4 - n^2b = n^2m^2 \implies n^2(n^2 - m^2) = n^2b \implies \boxed{b = n^2 - m^2}$

$\implies c = 2n^2 - (n^2 - m^2) = \boxed{n^2 + m^2}$

$\implies (a,b,c) = (2nm,n^2 - m^2, n^2 + m^2)$ and $(n,m) = 1$ , $n$ is odd and $m$ is even

and $n > m \implies (a,b,c)$ is a primitive pythagorean triple $\implies$

$\implies a + b + c = 2n^2 + 2nm = 2nm + 56n + 186 \implies$

$2(n^2 - 56n - 186) = 0 \implies 2(n + 3)(n - 31) = 0$ and $n > 0 \implies n = 31$

$\implies k = \lfloor{\dfrac{31}{2}\rfloor} = \boxed{15}$ .

Note: I used the following theorem below:

Let $n$ and $m$ be positive integers with $n > m$ and $a = n^2 - m^2, b = 2nm, c = n^2 + m^2$ .

The triple $(a,b,c)$ is a primitive pythagorean triple if and only if $gcf(m,n) = 1$ and $m$ and $n$ are not both odd.