Interesting fact about perfect squares

Hi, I found out something interesting about perfect squares and I would like to share it.

I don't know if anyone knew this, but I believe that most of you might knew this.

This is a quite easier way to find the square of a large number

Alright, let's start with two unknowns. I'll take xx and yy while y=x+1y = x+1

As we knew that x2=x×xx^2 = x\times x and y2=y×yy^2 = y\times y, we can change y2y^2 to be

y×y=y(x+1)=y×x+yy\times y = y(x+1) = y\times x + y

=x(x+1)+y=x×x+x+y=x(x+1) + y = x\times x + x + y

=x2+x+y=x^2 + x + y

y2=x2+x+yy^2 = x^2+x+y

Now, we have a simplified version of y2y^2

Let's try x=3x = 3 and y=4y = 4

42=32+3+44^2 = 3^2 + 3 + 4

16=9+3+416 = 9 + 3 + 4

And 16 is indeed equal to 9+3+49 + 3 + 4

So it worked! Yeah!

Now let's try a larger number, 501

If y=501,x=500y = 501, x = 500

5012=5002+500+501501^2 = 500^2 + 500 + 501

5012=250000+500+501501^2 = 250000 + 500 + 501

5012=251001501^2 = 251001

Try it in your calculator and you will find it true

Alright, that's all. Thanks everyone for viewing this.

Try this problem to see if you understand it

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Note by Daniel Lim
5 years, 9 months ago

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Why complicate the matter when you have got such a nice formula that you have been learning since "AGES"..... (a^ {2} - b^ {2} ) = (a-b)(a+b)

Priyansh Agrawal - 5 years, 9 months ago

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You can also think of it as this: Suppose x is 10 and y is 11. Imagine you have a square of 100 counters laid out in a 10 by 10 fashion. By adding 10 counters to the right of your grid, you get 110 counters in an 11x10 fashion. Then, add 11 counters to the bottom of the new grid and you get an 11 by 11 SQUARE grid. This shows that x^2+x+y=y^2. This also works for all consecutive numbers (you can try it yourself with counters.) Just another way to think of this method as a whole without using any algebra.

Yuxuan Seah - 5 years, 9 months ago

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That was the way I learnt it 0.o

Joshua Ong - 5 years, 5 months ago

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a^2- b^2 = (a+b)(a-b), an EASIER way!

Tigmanshu Anand - 4 years, 7 months ago

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This works is because (x1)2x2=2x1(x-1)^2-x^2=2x-1, and as we know, all odds can be written as the sum of two consecutive numbers. But this do save a little time. How about this?

For every two digit which has the form of a5 \overline {a5}, when they multiply with another number with the same form, then, a5×b5=wxyz \overline {a5} \times \overline {b5} = \overline{wxyz}, wx=a(a+1),yz=25\overline {wx} = a(a+1), \overline {yz} = 25.

敬全 钟 - 5 years, 9 months ago

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don't understand the a5b5a5 b5 part

Daniel Lim - 5 years, 9 months ago

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I want to clarify that a5\overline{a5} is not a×5a \times 5, but a5=10a+5\overline {a5} = 10a + 5.

敬全 钟 - 5 years, 9 months ago

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Well it isn't a simplification since we have to do 3 more operations instead of just one.

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(a + 1)^2 = a^2 + 2a + 1 = a^2 + a + (a + 1) just what you had above. Let us see a two digit number.
(10a + b)^2 = 100a^2 + b^2 + 20ab........... (10a - b)^2 = 100a^2 + b^2 - 20ab
Say 63^2 = (60+3)^2 = 3609 + 2018 = 3969 ...............67^ 2 = (70 - 3)^2 = 4909 - 2073 = 4489.
If b < 5 use (10a + b)^2, b > 5 use {(10 +1)a - (10 - b) }^2
If b=5, (10a + 5)^2 = 100
a*(a+1) +25

Niranjan Khanderia - 5 years, 4 months ago

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