Pythagorean triple?

For $n^{2} + (n + 1)^{2}$ to be a perfect square, there must exist a positive integer $m$ such that

$m^{2} = n^{2} + (n + 1)^{2} = 2n^{2} + 2n + 1$

$\Longrightarrow 2m^{2} = 4n^{2} + 4n + 1 + 1 \Longrightarrow 2m^{2} + (2n + 1)^{2} + 1 \Longrightarrow (2n + 1)^2 - 2m^{2} = -1$ .

This last equation is in the form of a negative Pell's equation, i.e., $x^{2} - dy^{2} = -1$ with $d$ a non-square integer. In this case $x = 2n + 1, y = m$ and $d = 2$ .

There are an infinite number of solutions to this equation, but the first few that are relevant to this problem are $(x,y) = (1,1), (7,5), (41,29), (239,169)$ , yielding values for $x = n$ of $0, 3, 20$ and $119$ . Since we are only looking for positive integers $n$ the sum comes out to $3 + 20 + 119 = \boxed{142}$ .

Comment: Since Pell's equations have been extensively studied I didn't go into details about how to obtain the first few relevant solutions. I will make note, though, of a recursive method of finding 'successive' solutions given a 'fundamental' solution. If $(x_{n}, y_{n})$ is a positive solution then $(x_{n+1}, y_{n+1})$ is given by $x_{n+1} = 3x_{n} + 4y_{n}$ and $y_{n+1} = 2x_{n} + 3y_{n}$ . So starting with a fundamental solution of $(1,1)$ , the subsequent solutions are $(7,5), (41, 29), (239,169), (1393,985), ....$ . This last pair yields a value for $n$ of $696 \gt 200$ , so neither this nor any subsequent pairs were relevant to this particular question.

The three such numbers are 3 , 20 and 119 . therefore the sum is 142

I used TI-83 PLUS and programed for
$x=0~~ ENTER\\ x~+~1~~ STO~~ x:~~~~\sqrt{x^2+(x+1)^2} \\\text{and pressed for 200 entries, and noted any integer answer..}\\n=3,~20~ and~ 119~~to~~get~sqares~of~~5,~29,~and~169~respectively.$