# Diophantine Equation Related To Triangles

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.

• You can say exactly the same. For a given number $k$ - the number of solutions of course. Therefore, it is necessary to change the condition. So all values were variable. Unless of course you want to get the formula. Commented Oct 2, 2014 at 5:54

## 2 Answers

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.

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

• Check again. In my opinion not correct. Commented Oct 2, 2014 at 6:52
• Sorry, I made a mistake. Is it correct now?
– user125368
Commented Oct 2, 2014 at 7:08
• Not correctly. You have the right part is always positive. And the left can be negative. This way is impossible equation to solve. Ask: $a,b,c$ -and already $k$ introduced as a multiplier. Commented Oct 2, 2014 at 7:45
• $p^2q/(2p+q) <= r^2 <= pq$ Now I think the solution is correct
– user125368
Commented Oct 2, 2014 at 8:05
• I think the answer should be edited. And to write more clearly. Commented Oct 2, 2014 at 8:08