Is there any natural number triangle whose inscribed circle's radius is $1$ except length $(a,b,c)=(3,4,5)$? 
Is there any natural number triangle that inscribed circle's radius is $1$ except length $(a,b,c)=(3,4,5)$? 

I found that there are no right triangle except $(3,4,5)$.

Thm. There are only one natural number right triangle that inscribed circle's radius is 1.

Proof)WLOG, $0<a<b<c$. Than $$a^2+b^2=c^2$$ $$a+b+c=ab$$. Because it is right triangle, $$a-1+b-1=c$$ .So, $$b=\frac{2(a-1)}{(a-2)}$$


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*If $a$ is even, $b$ is not a natural number.Because if $a=2k$ ,$b=\frac{2k-1}{k-1}=2+\frac{1}{k-1}$. Only integer solution in $k=2$.

*If $a$ is odd, $a=2k+1$ ,$a-1$is even but $a-2$ is odd. So $b$ is not integer except $a-2=1$.
I want to know general case about this. Such triangle is exist or not exist. I think It doesn't exist.

I think about it and I got one. Check It for me wrong or not.
We could make $$(a,b,c) =(x+y,y+z,z+x)$$ and by Heron's formula radius $$1= \frac{\sqrt{xyz}}{\sqrt{x+y+z}}$$ so $$x+y+z=xyz$$ and we knows only solution of this equation is $(x,y,z)=(3,2,1)$.

Say $a\leq b\leq c$.  If $s=(a+b+c)/2$ then $$r={S\over s}$$ where $S$ is area. By Heron formula we have
$$S=s \implies \boxed{(s-a)(s-b)(s-c)=s}$$
then $$s-a\mid s\implies s-a\mid a \implies a = k(s-a)$$
so $$k(b+c)=a(2+k)$$ Since $b+c\geq 2a$ we get $$2ka\leq a(2+k)\implies k\leq 2$$

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*$k= 1$ we have $b+c=3a$ and $s=2a$, so $$(2a-b)(2a-c)=2\implies -2a^2+bc=2$$
since $c= 2a-b$ we get
$$(2a-b)(b-a)=2$$
here we have 2 possibilities:

If $b-a = 1$ and $2a-b = 2$ we get $a=3$ and $b= 4$ so $c= 5$
If $b-a = 2$ and $2a-b = 1$ we get $a=3$ and $b= 5$ so $c= 4$

*

*$k=2$ then $b+c=2a$ so $a=b=c$, so $(s-a)^3=s$ so $a^3/8 = 3a/2$ so $a^2 = 12$ which is impossibile.

So the only such triangle has sides $3,4,5$.
