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

**Follow me to get more interesting questions and notes**

## Comments

Sort by:

TopNewestWhy 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 · 3 years, 8 months ago

Log in to reply

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 · 3 years, 8 months ago

Log in to reply

– Joshua Ong · 3 years, 4 months ago

That was the way I learnt it 0.oLog in to reply

a^2- b^2 = (a+b)(a-b), an EASIER way! – Tigmanshu Anand · 2 years, 6 months ago

Log in to reply

(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) +25 – Niranjan Khanderia · 3 years, 3 months agoIf b < 5 use (10a + b)^2, b > 5 use {(10 +1)a - (10 - b) }^2

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

Log in to reply

Well it isn't a simplification since we have to do 3 more operations instead of just one. – حكيم الفيلسوف الضائع · 3 years, 4 months ago

Log in to reply

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?

– 敬全 钟 · 3 years, 8 months agoLog in to reply

– Daniel Lim · 3 years, 8 months ago

don't understand the \(a5 b5\) partLog in to reply

– 敬全 钟 · 3 years, 8 months ago

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