TL;DR: The Trick.

Let’s first grok the trick just by pattern recognition.

Example: 31^2=

$$ \begin{aligned} 3\bold{1}^2 &= (\bold{3} * 10 + \bold{1}) \cr \bold{3} * (3\bold{1} + \bold{1}) &= 96\text{\textunderscore} \cr 1^2 &= \space\space\space\space 1\cr \hline &= 961 \end{aligned} $$

Example: 54^2=

$$ \begin{aligned} 5\bold{4}^2 &= (\bold{5} * 10 + \bold{4}) \cr \bold{5} * (5\bold{4} + \bold{4}) &= 290\text{\textunderscore} \cr 4^2 &= \space\space\space\space 16\cr \hline &= 2916 \end{aligned} $$

Example: 79^2=

$$ \begin{aligned} 7\bold{9}^2 &= (\bold{7} * 10 + \bold{9}) \cr \bold{7} * (7\bold{9} + \bold{9}) &= 616\text{\textunderscore} \cr 9^2 &= \space\space\space\space 81\cr \hline &= 6241 \end{aligned} $$

Example: 106^2=

$$ \begin{aligned} 10\bold{6}^2 &= (\bold{10} * 10 + \bold{6}) \cr \bold{10} * (10\bold{6} + \bold{6}) &= 1120\text{\textunderscore} \cr 6^2 &= \space\space\space\space\space\space 36\cr \hline &= 11236 \end{aligned} $$

Proof.

Let’s prove this result by using a similar approach that we took in “Proving the Squaring Numbers Ending in 5 Trick”.

$$ \begin{aligned} \text{Let: } a &= (10 * \bold{t} + \bold{n}) \space \text{where: }\bold{t} \geq 0, \bold{n} \geq 0 \cr \text{Then: } a^2 &= (10 * \bold{t} + \bold{n})^2 \cr &= (10 * \bold{t} + \bold{n}) (10 * \bold{t} + \bold{n}) \cr \end{aligned} $$

Let’s expand this, using FOIL. Our objective is to isolate a=(10t + n) in our simplification process:

$$ \begin{aligned} a^2 &= (10 * \bold{t} + \bold{n}) (10 * \bold{t} + \bold{n}) \cr &= \bold{10} * 10 * t * t + \bold{10} * t * n + \bold{10} * t * n + n^2 \cr &= 10 \Bigg[ 10 * \bold{t} * t + \bold{t} * n + \bold{t} * n \Bigg] + n^2 \cr &= 10 \Bigg[ t \bigg[ \bold{10 * t + n} + n \bigg] \Bigg] + n^2 \cr \text{Substitute: } a &= (10 * \bold{t} + \bold{n}) \cr &= 10 \Bigg[ t \bigg[ \bold{a} + n \bigg] \Bigg] + n^2 \cr \end{aligned} $$

That last result:

$$ a^2 = 10 \Bigg[ t \bigg[ \bold{a} + n \bigg] \Bigg] + n^2 $$

Proves our observation.

Example: 31^2=

Let’s consider our first example: 31^2 as a gut check –

$$ \begin{aligned} a &= 31 \cr &= (10 * \bold{3} + \bold{1}) \cr \text{Therefore: } \cr t &= 3 \cr n &= 1 \cr \text{Substitute. } \cr a^2 &= 10 \Bigg[ t \bigg[ \bold{a} + n \bigg] \Bigg] + n^2 \cr &= 10 \Bigg[ \bold{3} \bigg[ \bold{31} + \bold{1} \bigg] \Bigg] + (1)^2 \cr &= 10 \Bigg[ \bold{3} \bigg[ \bold{32} \bigg] \Bigg] + (1)^2 \cr &= 10 \Bigg[ \bold{96} \Bigg] + (1)^2 \cr &= 960 + 1 \cr &= 961 \cr \end{aligned} $$

Another way to look at this:

$$ \begin{aligned} 3\bold{1}^2 &= (\bold{3} * 10 + \bold{1}) \cr t \bigg[ \bold{a} + n \bigg] &= \bold{3} * (3\bold{1} + \bold{1}) = 96\text{\textunderscore} \cr n^2 &= 1^2 \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space = \space\space\space\space 1\cr \hline &= \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space 961 \end{aligned} $$

Reflections.

IMO, this representation of our a^2 identity:

$$ a^2 = 10 \Bigg[ t \bigg[ \bold{a} + n \bigg] \Bigg] + n^2 $$

is very nice to look at 😍😍.

Practically speaking, it may not be simple (nor would it be fast) to compute 7 * 88 (to square 79, for example) mentally. But knowing that the squaring of a number (a 2 digit by 2 digit operation) can be simplified to a simpler computation (a 1 digit by 2 digit operation, where the two factors being multiplied are determined by the actual number itself) is (to me at least) just super neat, man.

NOTE FOR MY FUTURE SELF: I ascertained this result by noticing that:

$$ \begin{aligned} 3^2 &= \space\space\space\space\bold{0}9 \cr 13^2 &= \space\space\bold{16}9 \cr 23^2 &= \space\space\bold{52}9 \cr 33^2 &= \bold{108}9 \end{aligned} $$

The terms that are not 9, I realized, were the sum of an arithmetic sequence where d=20 and the initial value was 16. In other words:

$$ \begin{aligned} \text{Let: } d &= 20 \cr 16 &= 0 + 16 + 0 * d \cr 52 &= 16 + 16 + 1 * d \cr 108 &= 52 + 16 + 2 * d \cr \end{aligned} $$

Because I knew d, the first term (16) and how many terms there would be in the sequence (3 terms if squaring 33, 2 terms if squaring 23, etc) I realized I could rewrite the numbers above as:

$$ \begin{aligned} \text{Sum of arithmetic seq: } S &= \dfrac{n}{2}\bigg(a_{0} + a_n \bigg) \cr 52 &= \dfrac{2}{2}\bigg(a_0 + a_1 \bigg) \cr &= \dfrac{2}{2}\bigg(a_0 + \big[ a_0 + 1 * d \big] \bigg) \cr &= \dfrac{2}{2}\bigg(16 + \big[ 16 + 1 * d \big] \bigg) \cr &= \dfrac{2}{2}\bigg(16 + \big[ 16 + 20 \big] \bigg) \cr &= \dfrac{2}{2}\bigg(16 + \big[ 36 \big] \bigg) \cr &= \bigg(16 + \big[ 36 \big] \bigg) \cr &= 2\bigg( \bold{26} \bigg) \cr 108 &= \dfrac{3}{2}\bigg(a_0 + a_2 \bigg) \cr &= \dfrac{3}{2}\bigg(a_0 + \big[ a_0 + 2 * d \big] \bigg) \cr &= \dfrac{3}{2}\bigg(16 + \big[ 16 + 2 * d \big] \bigg) \cr &= \dfrac{3}{2}\bigg(16 + \big[ 16 + 2 * \bold{20} \big] \bigg) \cr &= \dfrac{3}{2}\bigg(16 + \big[ 16 + 40 \big] \bigg) \cr &= \dfrac{3}{2}\bigg(16 + \big[ 56 \big] \bigg) \cr &= 3\bigg(8 + \big[ 28 \big] \bigg) \cr &= 3\bigg( \bold{36} \bigg) \cr &= \bold{108} \cr \end{aligned} $$

At this point, I noticed the pattern where:

$$ \begin{aligned} 52 &= 2\bigg( \bold{26} \bigg) \cr &= 2(23 + 3) \cr 108 &= 3\bigg( \bold{36} \bigg) \cr &= 3(33 + 3) \cr \end{aligned} $$

which strongly hinted that I might be able to isolate this form from expanding out a^2 where a=(10t+n).

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