@harsha: Thanks!

# Surprising Patterns in the Square Numbers (1, 4, 9, 16…)

**kalid**#42

@nelson: Awesome! In your case, it seems like x is the new number and x-1 is your original – another way to think about it :).

**kalid**#43

@John: Great explanation, I love seeing how other people think about it. I can already see the “a + b” version in my head, seeing how each side is growing (and one side is 1 unit longer than the other).

**crossbow**#44

Hi Kalid ,

I am preparing for a quantitative aptitude test and I always had the feeling of not having an aptitude for maths.After reading your blog, am feeling a little confident.Please give some insight on how to prepare for quantitative maths as I failed in my previous attempt for the above mentioned test.

**kalid**#45

@crossbow: Hi, my general advice would be to make sure you have an intuition for the concepts you have to learn, and then test your intuition by doing practice problems. Doing problems by themselves will only give you a mechanical understanding. It’s a very general question, but really try to develop your intuition for each topic as they come up. Good luck.

**000**#46

i found another intresting pattern i would like to show

a) 3 squared =9

b) 10 squared=1oo

c) 4 squared=16

d) 9 squared=81

e) 6 squared=36

<3 i love maths =D

**jenn**#49

To think about the cube method geometrically, I came up with the formula (x+1)^3 + x(2x+1).

(x+1)^3 represents the new top layer of the growing cube (assuming you are building up).

2x+1 is the new “thickness” around each of the existing layers. It is the same concept as you used for the growth of the area in the square example. There would be x layers where this new “thickness” is needed.

If you simplify (x+1)^3 +x(2x+1) it is equal to 3x^2+3x+1 which is the result when working with algebra or calculus.

I am interested if this is the way that other people thought of it, or if you set up a formula for geometric growth in a different way.

I really enjoyed the geometric, algebraic, and calculus connections in this post, but I agree that the geometric one is definitely the most intuitive, at least up to three dimensions.

**waqar**#50

I am looking for a specific pattern for squares of natural numbers

e.g, 4(power2)=1+2+3+4+3+2+1

**cole**#52

I found another instresting pattern I would like to show

1 1 after 0 +5

4 1 before 5 +5

9 1 before 10 +5

16 1 after 15 +10

25 On 25 +10

36 1 after 35 +15

49 1 before 50 +15

64 1 before 65 +15

Continue the same pattern, after before before after on, after before before after, on, forever and ever I found this pattern when playing with squares in my study hall. also it’s +5+5+5+10+10+15+15+15+20+20+25+25+25+30+30+35+35+35, ect.

**john**#54

I actually found the algebraic pattern one day while entertaining myself in a boring math lecture, and my mind almost exploded with excitement, except I did it a little bit differently.

although it is the same thing as x²+2x+1=(x+1)² I decided to give both numbers a variable, a and b, where b=a+1.

then it looks like a²+a+b=b²

ex 5²+5+6=36.

it seems more intuitive to me that instead of 2x+1, I saw it as a+b, and that the next number to be squared was already in the equation somewhere.

**dj1998**#55

Hey,

I figured that out myself though I really liked your visual tools.

You missed to write one thing though - The difference between 2 consecutive squares is odd and 2 consecutive differences are 2 consecutive odd numbers.

Helps in finding some squares.

**jk0**#57

Not sure if this is common knowledge though discovered it for myself earlier and can’t find any similar representations but yours:

x^2 = (x-1)*(x+1)+1

and after a few more calculations discovered:

x^2 = (x-y)*(x+y)+y^2

Don’t know how much use this is to anyone but thought I’d share it all the same

**ahmed**#58

I Have Been Thinking for that along time and Managed to find a relation from this relation that can get any Square Number

suppose we want to know 6^2 it will be equal to the distance from 1^2 to 6^2 + 1 so if we said x = 6 h = 6-1 = 5 there for 6^2 = 1 + h(2x+h) = 1 + 5 (2(1) + 5) = 36 so we can use this as a common formula for any number n^2 = 1+ h (2 + h) where h equals n-1 , and the same for cubes with the same method we can get a general formula which is 1+ h(3+3h+h^2) check for both it will work

Sry for my bad English