Part 1: the Basic Trick

Let’s begin with a very simplistic version of this trick.

Consider:

$$ \begin{aligned} 11 ^ 2 &= ? \cr &= 10 * 10 + 11 + 10 \cr &= 100 + 21 \cr &= 121 \end{aligned} $$

In words:

Suppose we have a number N to square. If we know the square of N-1, the square of N is simply (N-1)(N-1) plus N plus N-1.

How neat is that!

Let’s prove it:

$$ \begin{aligned} N^2 &= N * N \cr &= \big([N-1]+1\big)\big([N-1]+1\big) \cr &= [N-1]^2 + \big([N-1]+1\big) + \big([N-1]+1\big) + 1 \cr &= [N-1]^2 + \big(N\big) + \big([N-1]+1\big) + 1 \cr &= [N-1]^2 + \big(N\big) + \big(N\big) + 1 \cr &= [N-1]^2 + \bold{\big(N\big)} + \bold{\big(N + 1\big)} \cr \end{aligned} $$

Ta da!

While this is a great insight, the utilitiy of this approach breaks down if we want to know the square of some number N but we do not know the square of N-1.

For instance, what if we wanted to compute 14^2 but did not know 13^2? The question to ask now is:

Can this be generalized further such that we can compute the square of any number N = 10t * n (where t and n is a decomposed representation of the number, for instance if N=15, t is 1 and n is 5) provided we know the square of 10t (in this case, since t=1, 10t would just be 10)?

Part 2: the Second Trick

Suppose we wanted to find 14^2 but did not know 13^2 of hand (but obviously 10^2 is 100). Can we still find our answer?

From our obervation above:

$$ \begin{aligned} 14^2 &= 13^2 + 13 + 14 \cr &= \big(12^2 + 12 + 13 \big) + 13 + 14 \cr &= 12^2 + \big(12 + 13 \big) + \big(13 + 14 \big)\cr &= 11^2 + \big(11 + 12 \big) + \big(12 + 13 \big) + \big(13 + 14 \big)\cr &= 10^2 + \big(\bold{10} + 11 \big) + \big(\bold{11} + 12 \big) + \big(\bold{12} + 13 \big) + \big(\bold{13} + 14 \big)\cr &= 10^2 + \bold{\big(10 + 11 + 12 + 13\big)} + \big(11 + 12 + 13 + 14 \big) \cr \end{aligned} $$

This pattern looks interesting! Let’s generalize it a bit:

$$ \begin{aligned} \text{Let: } \cr t &= 1 \cr n &= 4 \cr a &= 10t + n = 14 \cr a - 1 &= 13 \cr \end{aligned} $$

From:

$$ \begin{aligned} 14^2 &= 10^2 + \bold{\big(10 + 11 + 12 + 13\big)} + \big(11 + 12 + 13 + 14 \big) \cr \end{aligned} $$

Notice that:

$$ \begin{aligned} \big(10 + 11 + 12 + 13\big) &= \sum_{10}^{13} x_i \cr \end{aligned} $$

So:

$$ \begin{aligned} \big(10 + 11 + 12 + 13\big) &= \cr &= \big(10t + \cdots + (a-1)\big) \cr &= \frac{4}{2}(10 + 13) \cr &= \frac{n}{2}(10t + \big[ (a-1) \big]) \cr \end{aligned} $$

And:

$$ \begin{aligned} \big(11 + 12 + 13 + 14\big) &= \sum_{11}^{14} x_i \cr \end{aligned} $$

So:

$$ \begin{aligned} \big(11 + 12 + 13 + 14\big) &= \cr &= \big(\big[ 10t + 1 \big] + \cdots + a\big) \cr &= \frac{4}{2}(11 + 14) \cr &= \frac{n}{2}(\big[ (10t + 1) \big]+ a) \cr \end{aligned} $$

Let’s substitute:

$$ \begin{aligned} 14^2 &= 10^2 + \bold{\big(10 + 11 + 12 + 13\big)} + \big(11 + 12 + 13 + 14 \big) \cr 14^2 &= 100t^2 + \frac{n}{2}\big[ 10t + (a-1) \big] + \frac{n}{2}\big[ (10t + 1) + a \big] \cr \end{aligned} $$

Simplify:

$$ \begin{aligned} 14^2 &= 100t^2 + \frac{n}{2}\big[ 10t + a \big] - \bold{\frac{n}{2}} + \frac{n}{2}\big[ (10t + a \big] + \bold{\frac{n}{2}}\cr \end{aligned} $$

Rearrange:

$$ \begin{aligned} 14^2 &= 100t^2 + \frac{n}{2}\big[ 10t + a \big] + \frac{n}{2}\big[ (10t + a \big] - \bold{\frac{n}{2}} + \bold{\frac{n}{2}}\cr \end{aligned} $$

Combine:

$$ \begin{aligned} 14^2 &= 100t^2 + \frac{n}{2}\big[ 10t + a \big] + \frac{n}{2}\big[ (10t + a \big]\cr \end{aligned} $$

And finally…

$$ \begin{aligned} (10t + n)^2 &= 100t^2 + n\big[ 10t + a \big]\cr \end{aligned} $$

Ta da!

Examples

Lete’s see if this works!

Example: 31^2=

$$ \begin{aligned} \text{Let: } \cr t &= 3 \cr n &= 1\cr a &= 10t + n = 31 \cr a - 1 &= 10 \cr \text{With: } \cr (10t + n)^2 &= 100t^2 + n\big[ 10t + a \big]\cr \text{Then: } \cr 31^2 &= 100(3)^2 + (1)\big[ 10(3) + 31 \big]\cr &= 100(3)^2 + \big[ 30 + 31 \big]\cr &= 900 + \big[ 61 \big]\cr &= 961 \cr \end{aligned} $$

Example: 39^2=

$$ \begin{aligned} \text{Let: } \cr t &= 3 \cr n &= 9\cr a &= 10t + n = 39 \cr a - 1 &= 38 \cr \text{With: } \cr (10t + n)^2 &= 100t^2 + n\big[ 10t + a \big]\cr \text{Then: } \cr 39^2 &= 100(3)^2 + (9)\big[ 10(3) + 39 \big]\cr &= 100(3)^2 + 9\big[ 30 + 39 \big]\cr &= 900 + 9\big[ 69 \big]\cr &= 900 + \big[ 630 - 9 \big]\cr &= 900 + \big[ 521 \big]\cr &= 1521 \cr \end{aligned} $$

^ Note, this isn’t as fun for numbers where n > 5. However, we can easily also solve this as:

$$ \begin{aligned} 39^2 &= 40^2 - 40 - 39\cr &= 1600 - 79\cr &= 1521\cr \end{aligned} $$

Reflections.

This one is pretty neat too, though I still contend that this solution,

$$ \begin{aligned} (10t + n)^2 &= 100t^2 + n\big[ 10t + a \big]\cr \end{aligned} $$

is not as pretty as:

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

(From: “Generic Trick for Squaring All Numbers”. )

But there you have it! Two approaches for squaring numbers without having to do the actual “math” of squaring stuff.

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