Squaring Numbers Faster than Calculator | Part-2 | x^2 - y^2 method

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In the previous post, we discussed, how to Square numbers faster than a calculator using the base method, but it required you to learn squares till 25, what if there existed a method where you don't require to learn squares till 25, and can still square numbers faster than a person can on a calculator? In this post, we will discuss a method, which just requires you to learn squares till 9! You just need to be a little fast in 2 digits by 1 digit multiplication, like 52*5.


The Recipe
The only prerequisite for this method is knowing squares till 9 by heart.

Let's start with very small numbers.

132
  1. Subtract the nearest multiple of 10 from the number, which in this case is 10. (13-10) = 3.
  2. Then, add the difference to the number. 13+3 = 16.
  3. Now, multiply the sum, with the nearest multiple of 10. 16*10 = 160.
  4. Now, square the difference, and add it to the product. 160+32 => 160+9 => 169
Let's try some more numbers.


222
  1. Subtract the nearest multiple of 10 from the number, which in this case is 20. (22-20) = 2.
  2. Then, add the difference to the number. 22+2 = 24.
  3. Now, multiply the sum, with the nearest multiple of 10. 24*20 = 480.
  4. Now, square the difference, and add it to the product. 480+22 => 480+22 => 484
372

  1. Subtract the nearest multiple of 10 from the number, which in this case is 40. (37-40) = 3.
  2. Then, add the difference to the number. 37+(-3) = 34.
  3. Now, multiply the sum, with the nearest multiple of 10. 34*40 = 1360.
  4. Now, square the difference, and add it to the product. 1360+32 => 1360+9 => 1369
We can also do that number like this,

372

  1. Subtract the nearest multiple of 10 from the number, we can also take the nearest multiple as 30. (37-30) =7.
  2. Then, add the difference to the number. 37+7 = 44.
  3. Now, multiply the sum, with the nearest multiple of 10. 44*30 = 1320.
  4. Now, square the difference, and add it to the product. 1320+72 => 1320+49 => 1369
Here are some more examples just to make everything more clear.

992
  1. (99-100) = 1
  2. 99+(-1)=98
  3. 98*100 = 9800
  4. 9800+12 = 9801
882
  1. (88-90) = -2
  2. 88+(-2)=86
  3. 86*90 = 7740
  4. 7740+22 = 7744
Now, just to make calculations a little bit more easy, you could have taken the nearest multiple as 100 also!


  2. 882

  1. (88-100) = -12
  2. 88+(-12)=76
  3. 76*100 = 7600
  4. 7600 + 122 = 7600+144 => 7744


How this trick works?
Let's say, we have a number, x. We have to find, its square.
To find: x2
x2 can be written as 
x2 -y2 + y2 .....(1)
x2-y2 can be written as (x+y)(x-y) .... (2)
Put (2) in (1)...
x2 = (x+y)(x-y) + y2
where y is the difference between the number and nearest multiple of ten.
Let's substitute some value of x.
Let x = 55,
552 = 552 + 52 - 52
=> (55+5)(55-5) + 52
=> (50)(60) + 25
=> 3000  + 25
=> 3025
Bonus
Whenever you want to multiply a number ending with the digit 5, you can just multiply the rest of the part with its successor, and that will become the left side of the answer, and right side of the answer will be 25. For eg, if you want to find the square of 65,
The answer will be in the form, 6*7/25 => 4225
If you want to why this works, just look at the proof above.
Here are some more examples, 
952 => 9*10/25 => 9025
452 => 4*5/25 => 2025
1052 => 10*11/25 > 11025
2252 => 22*23/25 => 50625


I hope this post was worth reading and if it was, then please don't forget to share this method with other people. Let me know your views about this method in the comments down below.

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