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 \(x\) and \(y\) while \(y = x+1\)

As we knew that \(x^2 = x\times x\) and \(y^2 = y\times y\), we can change \(y^2\) to be

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

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

\(=x^2 + x + y\)

\(y^2 = x^2+x+y\)

Now, we have a simplified version of \(y^2\)

Let's try \(x = 3\) and \(y = 4\)

\(4^2 = 3^2 + 3 + 4\)

\(16 = 9 + 3 + 4\)

And 16 is indeed equal to \(9 + 3 + 4\)

So it worked! Yeah!

Now let's try a larger number, 501

If \(y = 501, x = 500\)

\(501^2 = 500^2 + 500 + 501\)

\(501^2 = 250000 + 500 + 501\)

\(501^2 = 251001\)

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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## Comments

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

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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 + 20

ab........... (10a - b)^2 = 100a^2 + b^2 - 20abSay 63^2 = (60+3)^2 = 3609 + 20

18 = 3969 ...............67^ 2 = (70 - 3)^2 = 4909 - 2073 = 4489.a*(a+1) +25If b < 5 use (10a + b)^2, b > 5 use {(10 +1)a - (10 - b) }^2

If b=5, (10a + 5)^2 = 100

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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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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)

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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.

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That was the way I learnt it 0.o

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This works is because \((x-1)^2-x^2=2x-1\), 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?

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don't understand the \(a5 b5\) part

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

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