The sum of two consecutive triangular numbers is a square number.

Proof

The $n$ th Triangular Number is given by

T_{n} = \frac{n (n + 1)}{2}

Invoking the principle of mathematical induction,

\begin{aligned} T_{n} + T_{n + 1} & = \frac{1}{2} (n) (n + 1) + \frac{1}{2} (n + 1) (n + 2) \\ = \frac{1}{2} (n + 1) (n + (n + 2)) \\ = \frac{1}{2} (n + 1) (2 n + 2) \\ = (n + 1) (n + 1) \\ = (n + 1)^{2} \end{aligned}

Where the triangle numbers $T_{n}$ are

\documentclass{standalone}
\usepackage{tikz}
%
\newcommand{\dotTriangle}[1]{%
	\begin{tikzpicture}
		\foreach \n in {1,...,#1}{%
	        \foreach \k in {1,...,\n}{%
	            \node[circle,fill] at (\k-\n/2,-{\n*sqrt(3)/2}) {};
	        }
	    }
	    \pgfmathsetmacro\TriangleNumber{int((#1*(#1+1))/2)}
	    \node at (1/2,0) {\Large$T_{#1}=\TriangleNumber$};
    \end{tikzpicture}
}
%
\begin{document}
	\foreach \i in {1,...,4}{%
		\dotTriangle{\i} \hspace{1cm}
	}
\end{document}

Such that for $n = 1$ ,

T_{1} + T_{2} = 1 + 3 = 4 = 2^{2} = (1 + 1)^{2}

QED.