Diophantine Equation Related To Triangles

Question

a,b and c are the sides of a triangle and a, b, c are integers. I need to solve the following Diophantine equation for positive integral values of k.

$bc(b+c-a) = k^{2}(a+b+c)$

I think some parametric solutions may exist for this equation. I am unable to find them. Any help will be appreciated.

Answer 1

I'm playing around here to see what happens.

Rewrite $bc(b+c-a) = k^{2}(a+b+c)$ as $bc(b+c)-k^{2}(b+c) = abc+k^{2}a$ or $(b+c)(bc-k^2) = a(k^2+bc)$ or $a = \frac{(b+c)(bc-k^2)}{k^2+bc} = \frac{(b+c)(bc+k^2-2k^2)}{bc+k^2} =b+c- \frac{(b+c)(2k^2)}{bc+k^2} $.

This shows that $bc > k^2$ and $(bc+k^2) | 2k^2(b+c)$.

I'm not sure where to go from here - it's late and I'm tired, so I'll stop.

Answer 2

After a few calculation, I arrived at following solution :

$a = p(q^2+r^2), b = q(p^2+r^2), c = (p+q)(pq-r^2)$

$p^2+r^2$ must be whole square, because $k^2 = (q^2+r^2)(pq-r^2)^2$

$a+b+c=2pq(p+q)$ and $b+c-a=2p(pq-r^2)$

So $bc(b+c-a)=2pq(p+q)(p^2+r^2)(pq-r^2)^2$

Hence $k^2=(p^2+r^2)(pq-r^2)^2$