Proof without words: $1+8\times\text{triangular number}$ is an odd perfect square A recent question asked how to show that $8T_n+1$ is a perfect square if $T_n$ is a triangular number. This follows immediately from $T_n=\frac12 n(n+1)\implies 8T_n+1=4n^2+4n+1=(2n+1)^2$.
Can this be proven without words?
 A: The answer turns out to be quite simple, as @JMoravitz notes in his comment. In fact, Mathworld's page on triangular numbers includes precisely the right image:

A: $$
\begin{matrix}
\color{red}{\bullet} & \color{red}{\bullet} & \color{red}{\bullet} & \color{red}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{green}{\bullet} \\
\color{orange}{\bullet} & \color{red}{\bullet} & \color{red}{\bullet} & \color{red}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{green}{\bullet} & \color{green}{\bullet} \\
\color{orange}{\bullet} & \color{orange}{\bullet} & \color{red}{\bullet} & \color{red}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} & \color{green}{\bullet} & \color{green}{\bullet} & \color{green}{\bullet} \\
\color{orange}{\bullet} & \color{orange}{\bullet} & \color{orange}{\bullet} & \color{red}{\bullet} & \color{blue}{\bullet} & \color{green}{\bullet} & \color{green}{\bullet} & \color{green}{\bullet} & \color{green}{\bullet} \\
\color{orange}{\bullet} & \color{orange}{\bullet} & \color{orange}{\bullet} & \color{orange}{\bullet} & \color{black}{\circ} & \color{purple}{\bullet} & \color{purple}{\bullet} & \color{purple}{\bullet} & \color{purple}{\bullet} \\
\color{cyan}{\bullet} & \color{cyan}{\bullet} & \color{cyan}{\bullet} & \color{cyan}{\bullet} & \color{magenta}{\bullet} & \color{yellow}{\bullet} & \color{purple}{\bullet} & \color{purple}{\bullet} & \color{purple}{\bullet} \\
\color{cyan}{\bullet} & \color{cyan}{\bullet} & \color{cyan}{\bullet} & \color{magenta}{\bullet} & \color{magenta}{\bullet} & \color{yellow}{\bullet} & \color{yellow}{\bullet} & \color{purple}{\bullet} & \color{purple}{\bullet} \\
\color{cyan}{\bullet} & \color{cyan}{\bullet} & \color{magenta}{\bullet} & \color{magenta}{\bullet} & \color{magenta}{\bullet} & \color{yellow}{\bullet} & \color{yellow}{\bullet} & \color{yellow}{\bullet} & \color{purple}{\bullet} \\
\color{cyan}{\bullet} & \color{magenta}{\bullet} & \color{magenta}{\bullet} & \color{magenta}{\bullet} & \color{magenta}{\bullet} & \color{yellow}{\bullet} & \color{yellow}{\bullet} & \color{yellow}{\bullet} & \color{yellow}{\bullet}
\end{matrix}
$$
A: And you can convert
the proof without words
back into algebra
(which has the advantage of
showing that 
it is true
for any $n$):
$\begin{array}\\
8T_n
&=4(2T_n)\\
&=4(2(\frac{n(n+1}{2}))\\
&=4(n(n+1))\\
&=4n^2+4n\\
&=4n^2+4n+1-1\\
&=(2n+1)^2-1\\
\end{array}
$
A: Let k be a triangular number
$\frac{n(n+1)}{2}=k,n \in \mathbb{N}$
$\implies n^2+n=2k$
$\implies n^2+n-2k=0$
$\implies n=\frac{-1 \pm \sqrt{1+8k}}{2}$
$n=\frac{-1-\sqrt{1+8k}}{2}=-\frac{1+\sqrt{1+8k}}{2}$ [not possible]
$\implies n= \frac{-1 + \sqrt{1+8k}}{2}$
$n \in \mathbb{N} \implies (-1 \pm \sqrt{1+8k})| 2 , \implies \sqrt{1+8k} \in \mathbb{N} , \sqrt{1+8k} \neq 1$
Say
$\sqrt{1+8k}=l$
$\implies 1+8k=l^2$[proved]
by the way this was my first answer
