Find the smallest $n>3$ such that $(1+2+3+\cdots+n)^2$ is a perfect fourth power.
I know that $(1+2+3+\cdots+n)^2$ is the same as $(n(n+1)/2)^2$, by the formula for the sum of an arithmetic series. So I want to find the smallest $n>3$ such that $n(n+1)/2$ is a perfect square, since then its square will be a perfect fourth power. I can start by trying some small values of $n$ and see if they work. For $n=4$, I get $4(4+1)/2=10$, which is not a perfect square. For $n=5$, I get $5(5+1)/2=15$, which is also not a perfect square. For $n=6$, I get $6(6+1)/2=21$, which is still not a perfect square. For $n=7$, I get $7(7+1)/2=28$, which is also not a perfect square. For $n=8$, I get $8(8+1)/2=36$, which is a perfect square. So $n=8$ is the smallest $n>3$ that works, and $(1+2+3+\cdots+8)^2=1296=6^4$. 