A triangular number is a positive integer that can be expressed in the form $t_n=1+2+3+\cdots+n$, for some positive integer $n$. The three smallest triangular numbers that are also perfect squares are $t_1=1=1^2, t_8=36=6^2,$ and $t_{49}=1225=35^2$. What is the sum of the digits of the fifth smallest triangular number that is also a perfect square?

As in the solution of the problem, $t_n=m^2$ means $(2n+1)^2-8m^2=1$, whose solutions come from $(x,y)\mapsto(3x+8y,\ x+3y)$ starting at $(3,1)$: $(3,1),(17,6),(99,35),(577,204),(3363,1189),\ldots$

The fifth one gives $n=\frac{3363-1}2=1681$ and $t_{1681}=1189^2=1413721$.

Its digit sum is $1+4+1+3+7+2+1=\boxed{19}$.