The third square-triangular number

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A number is square-triangular if it is at the same time a perfect square and a triangular number (a triangular number has the form $T_k=\frac{k(k+1)}{2}$). The first two square-triangular numbers are $1$ and $36$. Find the third.

Solution

Look among the triangular numbers for those that are also squares. We have $T_1=1=1^2$ and $T_8=\frac{8\cdot 9}{2}=36=6^2$. Continuing: $T_{49} = \frac{49\cdot 50}{2} = 1225 = 35^2.$ The third square-triangular number is therefore $\boxed{1225}$ (it is both $35^2$ and the $49$-th triangular number).