Sum of Consecutive Squares

Problems involving the sum of two consecutive squares often appear on tests. A useful technique is to express the second number in terms of the first, i.e., $n^2 + (n+1)^2$. Expanding this gives $n^2 + (n^2 + 2n + 1) = 2n^2 + 2n + 1$.

Typically, one of the squares will be a number ending in 5, which is easy to compute. For example, $35^2 + 36^2$ can be solved as $2 \times 35^2 + 2 \times 35 + 1$. Since $35^2 = 1225$, this becomes $2 \times 1225 + 70 + 1 = 2450 + 71 = 2521$. This is much better than squaring both numbers separately and then adding.