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number theory senior
Problem
A is a positive integer that can be expressed in the form , for some positive integer . The three smallest triangular numbers that are also perfect squares are , , and . What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?
(A)
(B)
(C)
(D)
Solution
We have . If is a perfect square, then it can be written as , where is a positive integer. Thus, . Rearranging, we get , a Pell equation. So must be a truncation of the continued fraction for : \begin{eqnarray} 1+\frac12&=&\frac{2\cdot1+1}{2\cdot1}\\ 1+\frac1{2+\frac1{2+\frac12}}&=&\frac{2\cdot8+1}{2\cdot6}\\ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac12}}}}&=&\frac{2\cdot49+1}{2\cdot35}\\ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac12}}}}}}&=&\frac{2\cdot288+1}{2\cdot204} \end{eqnarray} Therefore, , so the answer is .
Final answer
D