Putnam 2015 and Ravi Substitution
Let $T$ be the set of all triples $(a,b,c)$ of positive integers for which there exist triangles with side lengths $a,b,c$. Express
$$\sum_{(a,b,c)\in T}\frac{2^a}{3^b5^c}$$
as a rational number in lowest terms.
One of the solutions is as follow:
Let $a=\frac{y+z}{2}, b=\frac{z+x}{2}, c=\frac{x+y}{2}.$
Then $x,y,z$ are either all even or all odd. Hence the original sum equals
$$\sum_{x,y,z>0 \ odd}\frac{2^{(y+z)/2}}{3^{(z+x)/2}5^{(x+y)/2}}+\sum_{x,y,z>0 \ even}\frac{2^{(y+z)/2}}{3^{(z+x)/2}5^{(x+y)/2}}$$.
Then after a substitution, we can unify the two sums and obtain a geometric series and we are done.
I have two questions.
First, why the substitution $a=\frac{y+z}{2}, b=\frac{z+x}{2}, c=\frac{x+y}{2}$ instead of the "ordinary" ravi substitution $a=y+z, b=z+x, c=x+y$? I know they are also equivalent conditions for a triangle to exist, but this is the first time I see this substitution (divided by 2). Is it possible to evaluate the sum using the ordinary ravi substitution?
Second, I don't understand why $x,y,z$ are either all odd or all even in order to generate all the triangles...
Thanks!
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