Mental Math: Squares
Squaring numbers (e.g. \(56^2 = 56 \times 56 = 3136\)) is a surprisingly useful thing to be able to calculate mentally. Fortunately, it is also one of the easiest tricks to do! With just a little practice, you should be faster than a calculator, for 2-digit numbers!
The trick is very simple, and relies on a basic algebra fact:
\[ x^2 ~~=~~ (x - c)(x + c) + c^2 \definecolor{colorlbl}{RGB}{128,192,255} \newcommand{\lbl}[1]{ {\small\color{colorlbl}#1} } \]Here's how you use it. Let's say you have a number you want to square, like for our example above \(x=56\).
Now, we choose a \(c\). We can choose any value we want! However, we want to pick the value of \(c\) so that the calculation is easier. In this case, I notice that if I choose \(c=6\), then \((x-c)=56-6=50\), which is a nice round number.
This also means that \((x+c)=56+6=62\). Multiply \(50\) from the previous step with \(62\) from this step: \(50 \times 62 = 3100\) (hint: to multiply by \(50\), multiply by \(100\) and divide by \(2\)).
Finally, we compute \(c^2\). Here, that's \(6^2=36\), which we know from our times-table. We add that in to get the final result: \(3100+36=3136\).
Altogether, the math looks like:
\begin{align*} 56^2 &= (56-6)(56+6) + 6^2\\ &= 50 \times 62 + 36\\ &= 3100 + 36\\ &= 3136 \end{align*}It may also be helpful to visualize what we're doing with a diagram. To square a number, we 'go up and down' the same small amount, multiply the results, and add the square of the small amount:
\begin{alignat*}{5} & & 62 & &\\ & \lbl{+6}\nearrow & & \searrow\lbl{\times} &\\ 56^2 & & & & 3100 \xrightarrow{\lbl{\large+6^2}} 3136\\ & \lbl{-6}\searrow & & \nearrow\lbl{\times} &\\ & & 50 & & \end{alignat*} \[ 56^2 = (56-6)(56+6) + 6^2 \]Here's a three-digit example: \(192^2\). Therefore, \(x=192\) and let's choose \(c=8\). We go up and down to get \(184\) and \(200\). Multiply those together to get \(36800\). Finally, add \(8^2=64\) to get our final answer \(192^2=36864\):
\begin{alignat*}{5} & & 200 & &\\ & \lbl{+8}\nearrow & & \searrow\lbl{\times} &\\ 192^2 & & & & 36800 \xrightarrow{\lbl{\large+8^2}} 36864\\ & \lbl{-8}\searrow & & \nearrow\lbl{\times} &\\ & & 184 & & \end{alignat*} \[ 192^2 = (192-8)(192+8) + 8^2 \]Notice that in the first example, I chose \(c\) so that \((x-c)\) was a nice number, whereas in the second example I chose \(c\) so that \((x+c)\) was a nice number. You should try to make at least one of them nice, but it doesn't matter which one.
Sometimes, you will get lucky and you can make both of them nice! When this happens, the calculation is particularly easy.
For example, suppose we have \(35^2\). Therefore, \(x=35\) and let's choose \(c=5\). We go up and down to get \(30\) and \(40\). These multiply together very easily to get \(1200\). Finally, you add \(5^2=25\) to get the final answer \(35^2=1225\).
Two-digit numbers ending in \(5\) will always work out this way. Try \(65^2\). Well, \(65\) is between \(60\) and \(70\), so multiply \(6 \times 7 = 42\) and tack a \(25\) on the end: \(4225\). Easy!
One beautiful aspect of the squaring numbers algorithm that is not often mentioned is that the final step at the end is itself another squaring operation! In the examples so far, it has just been a single-digit number from our times-table. But what if it's larger?
Let's say we want to square 219. The natural choice is \(c=19\). So, we go up and down to get \(200\) and \(238\), which we multiply to get \(47600\). However, for the last step, we have to add \(19^2\). You might not know what that is!
Fortunately, we now have a way to compute squares. Just apply our algorithm again, recursively! Start with \(19\), go up and down to get \(20\) and \(18\), multiply together to get \(360\) and add \(1^2\) to get \(19^2=361\).
Now, we can add \(361\) back in to \(47600\) to get our final answer of \(219^2 = 47961\):
\begin{alignat*}{5} & & 238 & &\\ & \lbl{+19}\nearrow & & \searrow\lbl{\times} &\\ 219^2 & & & & 47600 \xrightarrow{\lbl{\large+19^2}} 47961\\ & \lbl{-19}\searrow & & \nearrow\lbl{\times} &\\ & & 200 & & \end{alignat*} \begin{alignat*}{5} & & 20 & &\\ & \lbl{+1}\nearrow & & \searrow\lbl{\times} &\\ 19^2 & & & & 360 \xrightarrow{\lbl{\large+1^2}} 361\\ & \lbl{-1}\searrow & & \nearrow\lbl{\times} &\\ & & 18 & &\\ \end{alignat*} \begin{align*} 219^2 &= (219-19)(219+19) + 19^2\\ 19^2 &= (19-1)(19+1) + 1^2 \end{align*}Even more complicated problems might require two levels of substitution. For example, for \(12677^2\), you need \(2677^2\). For \(2677^2\), you need \(323^2\). For \(323^2\), you need \(23^2\). For \(23^2\) you need \(3^2\). And hopefully you know \(3^2\).