Ramanujan's house number

Statement

On one side of a street the houses are numbered consecutively $1,2,3,\dots,m$, with more than $20$ but fewer than $100$ houses. There is a house numbered $h$ such that the sum of the house numbers to its left equals the sum of those to its right. Find $h$ and the total number of houses $m$.

Solution

The condition is $1+2+\dots+(h-1)=(h+1)+\dots+m$, i.e. $T_{h-1}=T_m-T_h$, giving $T_m=T_{h-1}+T_h=h^2.$ So we need a triangular number $T_m$ that is also a perfect square $h^2$, with $20<m<100$. The suitable square-triangular number is $T_{49}=1225=35^2$, hence $m=\boxed{49},\qquad h=\boxed{35}.$ Check: $1+\dots+34=T_{34}=595$ and $36+\dots+49=T_{49}-T_{35}=1225-630=595$.