This approach my friends is pretty damn cool 😎.

Derivation

Let’s get straight to the derivation here as IMO that is the best way to grok this problem. Suppose we have some number - let’s say 82 - that we’d like to square.

Moreover, suppose there was a number close to 82 that we knew how to square easily. In this case, let’s go with 80 since \( 80^2 = \bold{6400} \) (and this is something we can easily compute mentally).

Now comes the fun part, let:

• \( \bold{b} = 80 \): the number that we can square easily
• \( \bold{c} = 82 \): the number that we wish to find the solution for

Also, lete there be some number \( a \) such that:

$$ a^2 + b^2 = c^2 $$

Ok, symbolically, we solve for \(a\) here easily enough:

$$ a^2 = c^2 - b^2 $$

By itself, this is not super useful. However, recall that the difference of two squares can be simplied such that:

$$ c^2 - b^2 = (c+b)(c-b) $$

Meaning:

$$ a^2 = (c+b)(c-b) $$

Let’s now substitute super quick with the example values we chose:

$$ \begin{aligned} a^2 &= (c+b)(c-b) \cr &= (82+80)(82-80) \cr &= (162)(2) \cr \bold{a^2} &= \bold{324} \end{aligned} $$

Ok, so what? Well, if we add this back to \( 80^2 \):

$$ \begin{aligned} a^2 + b^2 &= c^2 \cr a^2 &= 324 \cr b^2 &= 6400 \cr 324 + 6400 &= c^2 \cr 6724 &= (82)^2 \cr \end{aligned} $$

Tada! This approach is super neat because you can use any square you know as a starting point!

For instance, suppose we wanted to find \( 88^2 \) and we wanted to start with \( 90^2 \) since we can compute the square in our head. Then:

$$ \begin{aligned} a^2 + b^2 &= c^2 \cr c^2 &= (88)^2 \cr b^2 &= (90)^2 \cr &= 8100 \cr a^2 &= (c+b)(c-b) \cr &= (88+90)(88-90) \cr &= (178)(-2) \cr &= -356 \cr -356 + 8100 &= c^2 \cr 7744 &= (88)^2 \cr \end{aligned} $$

Ok, that one wasn’t a great example for mental math (but kudos to you if you can quickly (and accurately!) subtract 356 from 8100 in your head!) but overall this principle works great! Let’s try that same problem above but know our “known” starting point is \( 85^2 \) (which we can compute easily with the ending in 5 trick).

$$ \begin{aligned} a^2 + b^2 &= c^2 \cr c^2 &= (88)^2 \cr b^2 &= (85)^2 \cr &= 7225 \cr a^2 &= (c+b)(c-b) \cr &= (88+85)(88-85) \cr &= (173)(3) \cr &= 519 \cr 519 + 7225 &= c^2 \cr 7744 &= (88)^2 \cr \end{aligned} $$

(FWIW, \( 519 + 7225 \) may some scary at first to perform mentally but I swear it isn’t! Just add \( 520 \) to \( 7200 \) (that should be easy: \( 7720 \)) and then add back \( 25 \), so \( 7745 \) and then take away \( 1 \) which gets you \( 7744\) )

Quick Proof

Here’s a quick proof of the above. As we said, let there be some number \(c\) that we want to square. Additionally, let there be some number \(b\) that we already know how to square. Then, there is some number \( a \) such that:

$$ \begin{aligned} a^2 + b^2 &= c^2 \cr \end{aligned} $$

We can compute \(a\) “quickly” here by:

$$ \begin{aligned} a^2 + b^2 &= c^2 \cr a^2 &= c^2 - b^2 \cr a^2 &= (c+b)(c-b) \cr \end{aligned} $$

Finally, we can just substitute into our original equation to compute \(c\). This will always work because:

$$ \begin{aligned} a^2 + b^2 &= c^2 \cr a^2 &= (c+b)(c-b) \cr (c+b)(c-b) + b^2 &= c^2 \cr c^2 - b^2 + b^2 &= c^2 \cr c^2 &= c^2 \cr \end{aligned} $$

^ this might have been obvious but wanted to share out here to close the loop.

Recap

Ok so let’s recap this - given some number \( n \) you want to square:

1. Find a number close by to \( n \) that you know how to square in your head (either because you have it memorized or because you have a trick that makes it significantly easier). Let’s call this number \( \bold{k} \).
2. Compute \( \big[ n+k \big]\big[ n-k \big] \)
3. Add the result from (2) to \( \bold{k}^2 \)
4. ???
5. Profit!

I really like this one!

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