A Simple P.P.T Problem.

$a = \dfrac{1}{n}(b + c)$ and $a + b + c = m \implies b + c = m - a \implies$

$na = m - a \implies (n + 1)a = m \implies a = \dfrac{m}{n + 1} \implies b + c = \dfrac{mn}{n + 1}$

$\implies c = \dfrac{mn}{n + 1} - b \implies (\dfrac{mn}{n + 1} - b)^2 = m^2 + (n + 1)^2b^2 \implies$

$(mn - (n + 1)b)^2 = m^2 + (n + 1)^2b^2 \implies m^2n^2 - 2mn(n + 1)b + (n + 1)^2b^2 =$

$m^2 + (n + 1)^2b^2 \implies m^2(n^2 - 1) = 2mn(n + 1)b \implies$

$b = \dfrac{m^2(n^2 - 1)}{2mn(n + 1)} = \dfrac{m(n - 1)}{2n} \implies c = \dfrac{m(n^2 + 1)}{2n(n + 1)}$

$\implies (a,b,c) = (\dfrac{m}{n + 1}, \dfrac{m(n - 1)}{2n}, \dfrac{m(n^2 + 1)}{2n(n + 1)})$

Multiplying each side by $\dfrac{2n(n + 1)}{m}$ we obtain $(a,b,c) \sim (2n, n^2 - 1,n^2 + 1) =$

$(a^{*}, b^{*},c^{*})$ and $(n,1) = 1$ and $n$ is even $\implies (a^{*}, b^{*},c^{*})$ is a primitive pythagorean triple

$\implies a^{*} + b^{*} + c^{*} = 2n^2 + 2n = 10n - 8 \implies 2(n - 2)^2 = 0 \implies \boxed{n = 2}$ .

We are told that the perimeter of the original triangle is $m$ . When this triangle is scaled by a factor of $\frac{2n(n + 1)}{m}$ , the perimeter becomes $2n(n + 1)$ . So $2n(n + 1) = 10n - 8.$ The solution to this equation is $n = 2$ .