<s>[INST] Find possible number of triangles with integer sides for a given inradius inradius = $5$
possible triangle sides
$(25, 20, 15)$
$(37, 35, 12)
(39, 28, 17)$
...
Find formula to find other possible sides of triangles.
 [/INST] Let $a$, $b$, $c$ be the integer lengths of the triangle sides, and $r$ the inradius, which I suppose to be integer too. It is well known that $r(a+b+c)$ is twice the area of triangle, so by squaring Heron's formula we have:
$$
r^2(a+b+c)={1\over4}(a+b-c)(a-b+c)(-a+b+c).
$$
The three factors in parentheses on the right hand side of that equation must be all even or all odd, because the sum of any two of them is twice the length of a side. But their product must be divisible by $4$, so they are all even numbers and we can set:
$$
2x=(a+b-c),\quad 2y=(a-b+c),\quad 2z=(-a+b+c).
$$
After substitution, the equation becomes
$$
r^2(x+y+z)=xyz,
$$
where $x$, $y$, $z$ are positive integers and $a=x+y$, $b=x+z$, $c=y+z$. We can solve for $z$ to get:
$$
z={r^2(x+y)\over xy-r^2}.
$$
All solutions can be obtained from that formula, by choosing $x$ and $y$ such that $xy>r^2$ and $z$ is integer. If we suppose, without loss of generality, that $z\ge x$ and $z\ge y$, then $2z\ge x+y$, which entails $xy\le 3r^2$. So we must check only a limited number of cases. 
A complete search in the case $r=5$ can be readily carried out and the only solutions for $(a,b,c)$ are the following:
(11, 60, 61),   (12, 35, 37),   (12, 153, 159), (13, 68, 75),  (15, 20, 25), 
(15, 377, 388), (17, 28, 39),   (17, 87, 100),  (27, 29, 52),  (27, 676, 701), 
(28, 351, 377), (31, 156, 185), (36, 91, 125),  (39, 76, 113), (51, 52, 101).

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