Squaring a Number in fewer steps

I will show you how to reduce the number of steps it takes to square a number, called Johnson's Method of Speed Squaring

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[Edit]
The method you learned in middle school

First we will go over the method you learned in school.  Let's square the number 543

First we write the number twice, one on top of the other like this


 543
          x  543
   ------

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Then we do the following multiplications 3x3 is 9, 4x3 is 12 carry the 1, 5x3 is 15 plus the carried one is 16.  Our result so far looks like this


  543
          x   543
   ------
   1629  

[Edit]
Now we do the following multiplications 3x4 is 12 carry the 1, 4x4 is 16 plus the carried one is 17 carry the 1, 4x5 is 20 plus the carried 1 is 21.  Our result so far looks like this


  543
          x   543
   --------
   1629  
  2172  

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Now we do the following multiplications 5x3 is 15 carry the 1, 5x4 is 20 plus the carried 1 is 21 carry the 2, 5x5 is 25 plus the carried two is 27.  Our result so far looks like this


  543
          x   543
   ---------
    1629  
   2172  
  2715  

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Finally we add all these columns up.  Bring down the 9 in the ones column.  2 + 2 is 4 for the tens column.  6+7+5 is 18 carry the 1 so we have an 8 in the hundreds column, 1+1+1 is 3 plus the carried one give us a 4 in the thousands column, 2+7 is 9 in the ten-thousands column, and bring down the 2 in the hundred-thousands column


  543
          x   543
   ---------
    1629  
   2172  
  2710  
  ----------
 
  294849  


[Edit]
Now I'll show you the faster method

First write out the number with zeroes between each digit like this

        050403

Now under each number writes it's square, so under 05 we write 25, under 04 we write 16 and under 03 we write 09 like this

    050403
    251609

Now in the spaces between each of those multiplications, we will multiple each digit by each other digit and then by two, and place the result, with the ones position starting at the midpoint between the digits.  Hold on, I know that makes little sense yet, let's take an example.  In the first line 050403 the midpoint between the "5" and the "4" is the zero between them.  So when we do our new calculation 5x4x2 which is 40, we will place the 0 at the point, and the 4 obviously in front of it like this.

    050403
    251609
     40

You see how the 40 is shifted so that it's first digit "0" is on the midpoint line between the "5" and the "4".  Let's do another example.  When we multiple 4x3x2 which is 24, we will place the "4" at the midpoint between the 4 and the 3 so it will go at the point where the 0 between the "4" and the "3" is above it. Like this

    050403
    251609
     4024

You see how the "24" goes below the "4" and the "0", the "0" in the first line, being the midpoint between the "4" and the "3".  Now let's do the last, slightly more complicated one, which is to multiple the "5" by the "3".  5x3x2 is which 30.  The midpoint between the "5" and the "3" is the "4", it is two positions to the right of the "5" and two positions to the left of the "3", so it is the midpoint between them.  So our new result "30" which start, that is, the zero will be place directly under the "4" like this

    050403
    251609
     4024
      30

Do you see how the "0" in "30" is directly under the "4" in the first line?

Now the final step, you add straight down like this

    050403
    251609
     4024
       30
-----------------
294849

And voila we get the exact same answer, as we did the longer way.

Now let's compare the amount of effort involved.

In the old fashioned method we required:
    three multiplications, one carry, one additions, writing four digits
    three multiplications, two carries, two additions, writing four digits
    three multiplications, two carries, two additions, writing four digits
    four additions and two bring-downs, writing six digits

In the new method we require:
    three multiplications, writing six digits
    three multiplications, three doublings, writing six digits
    four additions and two bring-downs, writing six digits

Most people find doubling a number to be very trivial compared to multiplication, which is why I call it by a different name here.

What this new method avoids completely is any type of carry and any addition related to the carry.  In addition, it makes three random multiplications into the much simpler "doubling".

This is Johnson's Method of Speed Squaring.