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Find number of triangles with integral sides and side lengths less than or equal to $2n$. I approached this method by recursion.

Say $A_{2n}\ $is the number of triangles with integral sides and side lengths less than or equal to $2n$. $A_{2n-1}\ $ is the number of triangles with integral sides and side lengths less than or equal to $2n-1$ .

So $A_{2n}=A_{2n-1}+\left( \text{number of triangles having at least one side equal to} 2n \right) .$

How to count the number of triangles having at least one side equal to $2n$?

Also, is there any other better method to this other than generalization as well?

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  • $\begingroup$ Perhaps it would be easier to count up instead of counting down... $\endgroup$ – abiessu Dec 17 '18 at 12:54
  • $\begingroup$ Welcome to Math.SE! Please use MathJax. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 17 '18 at 12:55
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Number of triangles having at least one side equal to $2n$ is the number of couple $(a,b)$ with $a\ge b$ and $a+b > 2n$. That is, $$\begin{aligned} \sum^{2n}_{b=1}\sum^{2n}_{a = \max(b,2n + 1-b)} 1 &= \sum^{2n}_{b=1} (2n+1 - \max(b,2n+1-b))\\ &=\sum^{2n}_{b=n+1} ((2n+1) - b) + \sum^{n}_{b=1} (2n+1 - (2n+1-b))\\ &=\sum^{1}_{c=n}c + \sum^{n}_{b=1} b\hspace{5em}(\text{where } c = 2n+1-b)\\ &= n(n+1) \end{aligned}$$ In order to argue recursively, you would need to compute $A_{2n-1}$ in similar fashion as well.

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