Triangle numbers that are squares of triangle numbers. What are the triangle numbers the are squares of other triangle numbers? I have found $1^2=1$ and $6^2=36$, but other than these examples I can't find any other triangle numbers that are squares of other triangle numbers, and I used a program to check this idea into the thousands.
Finding triangle numbers that are squares of other triangle numbers corresponds to finding integers $n$ and $k$ such that $n(n+1)/2=[k(k+1)/2]^2$, or such that 

$$2n(n+1)=k^2(k+1)^2 .$$

I believe the only positive integer solutions to this equation are $(1,1)$ and $(3,8)$, but I don't know how to prove it. Are there any others?

Note that $n$ and $n+1$ are coprime, as are $k$ and $k+1$.  We can have both sides zero if $n=0,-1, k=0,-1$.  Otherwise we must have either $n=k^2,2(n+1)=(k+1)^2$ or $n+1=k^2,2n=(k+1)^2$.  The first gives 
$$n=k^2\\2(n+1)=(k+1)^2\\2k^2+2=k^2+2k+1\\k^2-2k+1=0\\k=1\\n=1$$ while the second gives $$n+1=k^2\\2n=(k+1)^2\\2k^2-2=k^2+2k+1\\k^2-2k-3=0\\k=-1,3\\n=0,8$$
and you have found all the positive solutions.
