Why do consecutive triangular numbers in pairs like $6$ and $10$ always add up to a perfect square? I was a bit surprised by this when I thought of it.  Look here: $$001, 003, 006, 010, 015, 021, 028, 036, 045, 055$$$$004, 009, 016, 025, 036, 049, 064, 081, 100$$As you just saw, $15$ and $21$ add to make $36$, a perfect square, and $36$ and $45$ add to make $81$, another perfect square.  It seems that all consecutive triangular numbers in pairs always add to make perfect squares.  Why does this remarkable thing even happen?
 A: The formula for $T_n$ is
$$T_n = \frac{n(n+1)}2$$
Thus
$$\begin{align*}
T_n + T_{n+1} & = \frac{n(n+1)}2 + \frac{(n+1)(n+2)}{2}\\
& = \frac{n^2 + n + n^2 + 3n + 2}2\\
& = \frac{2n^2 + 4n + 2}2\\
& = n^2 + 2n + 1 = (n+1)^2
\end{align*}$$
Another way to see it
$T_n$ counts the number of dots in a $(n+1)\times(n+1)$ square wich are below the diagonal; $T_{n+1}$ counts these plus the diagonal. Equivalently $T_{n+1}$ counts the dots above or on the diagonal of a $(n+1)\times(n+1)$ square. Evidently, their sum must thus be the number of dots in the entire square, $(n+1)^2$.
$$\newcommand{\r}{{\color{red}\cdot}}\newcommand{\b}{{\color{blue}\cdot}}
\pmatrix{\b&\b&\b&\b\\
\r&\b&\b&\b\\
\r&\r&\b&\b\\
\r&\r&\r&\b} T_{\color{red}3}+T_{\color{blue}4}
$$
A: The $n^\text{th}$ triangular number is the sum of the first $n$ natural numbers, with the empty sum $T_0 = 0$. The $n^\text{th}$ perfect square is the sum of the first $n$ odd natural numbers.
\begin{align}
T_n &= 0 + 1 + 2 + \cdots + n\\\\
T_{n+1}&= 1 + 2 + 3 + \cdots + n+1 \\\\
T_n+T_{n+1}& = 1 + 3 + 5 + \cdots +2n+1\\\\
\end{align}
Since $2n+1$ is the $(n+1)^\text{th}$ odd natural number, we have 
$$T_n+T_{n+1}=(n+1)^2$$
