Triangle and Square Sequence

We are looking for Diophantine solutions to the equation $2m^2=n(n+1)$ . The factorization of $2m^2$ has a 2 and other primes that come in pairs. $n$ and $n+1$ do not share any factors, so one has to be a perfect square, and the other twice a perfect square. We now have two cases to consider. First Case: $n=2b^2$ and $n+1=a^2$ . For every $(a,b)$ that is a solution, $(3a+4b,2a+3b)$ is also another solution. It is straightforward to verify that $(3a+4b)^2=2(2a+3b)^2+1$ given that $a^2=2b^2+1$ . Rewriting this in terms of $n$ gives for every solution $n$ , $17n+8+12\sqrt{2n^2+2n}$ is also a solution. This will create a strictly increasing sequence of solutions starting with $n=0$ . We know that every decreasing sequence (the same process in reverse) must go to 0, because 8 goes to 0, so if another sequence exists, it must go somewhere between 0 and 8, but none of those options work. We can use the recursive formula $n_{k+1}=17n_k+8+12\sqrt{2n_k^2+2n_k}$ to generate every solution from 0. Second Case: $n=a^2$ and $n+1=2b^2$ This gives $a^2=2b^2-1$ . If $(a,b)$ is a solution, then so is $(3a+4b,2a+3b)$ It is straightforward to verify that $(3a+4b)^2=2(2a+3b)^2-1$ given that $a^2=2b^2-1$ . Rewriting this in terms of $n$ gives for every solution $n$ , $17n+8+12\sqrt{2n^2+2n}$ is also a solution. There are no solutions between $n=1$ and $n=49$ , so every possible solution is generated by the recursive formula $n_{k+1}=17n_k+8+12\sqrt{2n_k^2+2n_k}$ starting with $n=1$ . The next highest n for either of these lists is $n=49$ which gives an answer of 1225.

For a natural number p, p(2p+1) or p(2p-1) must be a perfect square. Since p and 2p+1 , as well as p and 2p-1 are relative primes, therefore both p and 2p+1 or both p and 2p-1 must be perfect squares. Such values of p are 1, 4, 25, ... Therefore the 3rd. member is 25(50-1) or 25*49 or 1225

The triangular numbers are given by n(n + 1)/2. The squares are given by k^2. Hence we seek solutions to n(n + 1)/2 = k^2, or n^2 + n = 2k^2. Multiplying by 4 and adding 1, we have (2n + 1)^2 = 8k^2 + 1. Defining x = 2n + 1 and y = 2k, the equation can be written as x^2 - 2k^2 = 1. This is Pell's equation which has a recursive solution whose first three iterates are: x 1 = 3, y 1 =2, x 2 =17, y 2 =12, and x 3 =99, y 2 = 70, from which we get k = 35, n = 49. n(n+1)/2 = 1225, K^2 = 1225. I didn't want to use up the space, but in case anyone is interested, the recursive formulae are: x 1 = 3, y 1 =2, x n+1 = (x 1)(x n) + 2(y 1)(y n) and y n+1 = (x 1)(y n) + (y 1)(x n). Ed Gray