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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.

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  • $\begingroup$ 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. $\endgroup$
    – individ
    Oct 2, 2014 at 5:54

2 Answers 2

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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.

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

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  • $\begingroup$ Check again. In my opinion not correct. $\endgroup$
    – individ
    Oct 2, 2014 at 6:52
  • $\begingroup$ Sorry, I made a mistake. Is it correct now? $\endgroup$
    – user125368
    Oct 2, 2014 at 7:08
  • $\begingroup$ 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. $\endgroup$
    – individ
    Oct 2, 2014 at 7:45
  • $\begingroup$ $p^2q/(2p+q) <= r^2 <= pq$ Now I think the solution is correct $\endgroup$
    – user125368
    Oct 2, 2014 at 8:05
  • $\begingroup$ I think the answer should be edited. And to write more clearly. $\endgroup$
    – individ
    Oct 2, 2014 at 8:08

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