#math/mental
For **squaring two-digit integers** we leverage the following $n^{2}=(n+d)\times(n-d)+d^{2}$. Consider $13^{2}$.
| Numbers that add to 26 | Distance from 13 | Their product | Product's difference from 169 |
| :--------------------: | :--------------: | :-----------: | :---------------------------: |
| 13, 13 | 0 | 169 | 0 |
| 12, 14 | 1 | 168 | 1 |
| 11, 15 | 2 | 165 | 4 |
| 10, 16 | 3 | 160 | 9 |
| 9, 17 | 4 | 153 | 16 |
| 8, 18 | 5 | 144 | 25 |
The product's difference from 169 is exactly the square of the distance ($0^2, 1^2, 2^2, \dots$). Then the problem can be simplified to
```tikz
\begin{document}
\begin{tikzpicture}[>=stealth]
\node (start) at (0,0) {$13^2$};
\node (top) at (2.2,1) {$16$};
\node (bot) at (2.2,-1) {$10$};
\node (end) at (5.6,0) {$160 + 3^2 = 169$};
\draw[->] (start) -- node[above,sloped] {$+3$} (top);
\draw[->] (start) -- node[below,sloped] {$-3$} (bot);
\draw[->] (top) -- (end);
\draw[->] (bot) -- (end);
\end{tikzpicture}
\end{document}
```
Rounding to the nearest 10 is generally the best idea. When **squaring a number ending in 5** calculate first digit times one digit above, then append 25. Consider $\textcolor{#1f9d57}{3}5^{2}$. Then $\textcolor{#1f9d57}{3}\times \textcolor{#1f9d57}{4}=12 \xrightarrow{\text{append 25}} 12\textcolor{#1f9d57}{25}$ .