Sum is Triangular

Write $\frac{n(n+1)}2 = (2k-1)^2 + (2k+1)^2$ . WLOG $k \ge 0$ . After multiplying by $8$ and adding $1$ to both sides, we get $(2n+1)^2 = (8k)^2 + 17$ So $(2n+1+8k)(2n+1-8k) = 17$ , and the only way to write it this way is if the first factor is $17$ and the second one is $1$ . So there is exactly one positive value of $n$ that works, namely $n = 4$ . So the answer is $\fbox{10}$ .