STARS wrote: Let $ a_n = 1 + 2 + ... + n$ for every $ n \ge 1$; the numbers an are called triangular. Prove that if $ 2a_m = a_n$ then $ a_{2m - n}$ is a perfect square. $ a_{n}=\frac{n(n+1)}{2}$-----> $ 2a_m = a_n$-----> $ m(m+1)=\frac{n(n+1)}{2}$---> $ 2m-n=n^{2}-2m^{2}$----> $ a_{2m-n}=\frac{(2m-n)(2m-n+1)}{2}=\frac{(2m-n)^{2}+2m-n}{2}=\frac{(2m-n)^{2}+n^{2}-2m^{2}}{2}=\frac{2(m-n)^{2}}{2}=(m-n)^{2}$