# Why are the differences between consecutive squares equal to the sequence of odd numbers?

I was playing around with the squares and saw an interesting pattern in their differences.

$0^2 = 0$

      + 1


$1^2 = 1$

      + 3


$2^2 = 4$

      + 5


$3^2 = 9$

      + 7


$4^2 = 16$

      + 9


$5^2 = 25$

      + 11


$6^2 = 36$

      etc.


(Also, in a very related question, which major Math Research Journal should I contact to publish my groundbreaking find in?)

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Could you send it to a social science journal instead? :-) –  Simon Nickerson Jul 20 '10 at 21:42
You can also reverse your observation (which as Dan explains is very easy to show) to get the slightly harder result that 1+3+5+...+(2n-1) = n^2. –  Noah Snyder Jul 20 '10 at 21:42
The comment about publishing in a major math research journal is presumably a joke, but it seems like some of the answers think it was meant seriously.... –  Nate Eldredge Feb 4 '13 at 22:14

Dan's algebraic justification is correct, but you may get more intuition about why this is happening from the above picture. Each time you want to enlarge the square by one unit, you have to add an extra row, an extra column, and one more square to fill in the corner. These correspond directly to the $n+n+1=2n+1$ that Dan mentioned. And of course, $2n+1$ is how odd numbers look.

Looking at it from the other direction, you can use the same idea to convince yourself of Noah's claim that $1+3+5+\dots+(2n-1) = n^2$. Imagine the left-hand side as representing the upper sequence of green and orange squares. You add on first 1 tile, then 3, then 5, and so on to construct larger and larger squares. At each step, you are adding one row (n) and one column (another n), then removing that one tile in the top right corner where they overlap (for a total of 2n-1).

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A nice thing to notice.

It's basically because

Difference between two consecutive square i.e:

$(n + 1) ^ 2 - n ^ 2 = 2n + 1$

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+1: And this would be the induction step in a formal proof, after noting the initial case being that the sum of the first 1 odd numbers is 1 ^2. –  donroby Jul 21 '10 at 1:17

The easiest way to I think about the difference between consecutive powers ( squared cubed or whatever) is using binomial theorem.

For example :

$(n+1)^4 - n^4 = n^4 + 4n^3 + 6n^2 + 4n + 1 - n^4 = 4n^3 + 6n^2 + 4n + 1$, with the binomial coefficients found by Whatever method you prefer.

This correctly predicts e.g $2^4-1^4=15$.
And when used on squares gives $2n+1$.

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Don't make this a paper. This has been done over the centuries by so many brilliant mathematicians.

Reading Gauss, Newton, Hardy, Littlewood, Ramanujam and so many other mathematicians over the years will tell you that everything that you imagine has already been done.

You have to read many papers for many years, understand them in depth to come up with new observations using those publications as your base.

But hey... don't stop finding stuff by yourself. It is a good thing to do this kind of stuff.

regarding difference between consecutive cubes, I am more interested in explaining to myself why $d(x^3)/dx = 3x^2$ whereas for natural numbers $(n+1)^3 -n^3 = 3n^2 + 3n + 1$ all I can think of is

$lim (3n^2 + 3n + 1)$ =~ $3n^2$ as $n^2$ dwarfs $3n$. $n\rightarrow \inf$

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This is one of the first questions on this site. I doubt the OP actually wanted to publish this. –  Asaf Karagila Oct 19 '12 at 12:46
'Everything you imagine has already been done'. While I understand where the sentiment is coming from, I cannot imagine a worse statement to greet someone coming into mathematics with. –  Steven Stadnicki Feb 4 '13 at 22:14

o.k. you see the difference between the squares, but you dont understand why. I also have noticed this differnce, but i took it a step further when i came up with equation (a+1)^2- (a)^2 + 2 + (a+1)^2 = (a+2)^2. i asked my question which you can see in the section tagged "consecutive squares". which was, "if it was part of a bigger equation but found out it wasn't". none the less it took the differences of a certain square and corelated it to the next consecutive square (such as the fact that 1^2=1, and 2^2=4 and the difference between them, as shown in the chart above, is 3 sure, great! but what of it ?), well if you add 2 to that difference than add it back to the (2^2, or 4) the larger consecutive square. Then you get 9 which is the next cosecutive square, (3^2). That is the concept of the equation above, technically i didnt explain that specific aspect of the difference of consecutive squares, but i did explain the general correlation between consecutive squares.

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o.h. by the way this was definitiley not discovered by you and as i said earlier this is only the begining to my real, complete equation, i didn't get any realization for my equation so if you think your going to get any thing for this concept than your mistaken. i came up with this simple concept years ago and then took it farther and made it a actual equation this is nothing more than a base to the real equation that i came up with. I credit none of my discovery to you, if they didn't even except what i came up with. there just going to laugh at you. –  user7911 Mar 7 '11 at 16:03

to extend this idea, there is something i came up with in high-school (now i suck at even elementary math)... I'm sure someone had come up with this much earlier ..

"the Nth difference between Nth consecutive powers is equal to N!"

for N=2
1  4  9  16  <- Nth powers
3  5  7     <- 1st Difference
2  2       <- Nth Difference

for N=3

1  8  27  64  125
7  19  37  61
12  18  24
6   6

for N=4

1  16  81  256  625  1296  2401
15  65  175  369  671
50  110  194  302
60   84   108
24    24
`

this can also be used to calculate the next number in the series eg: 7^4 = 24+108+302+671+1296 = 2401

i wish i could remember the proof too ... Yikes

so for N=2 difference is 2 hence leading to a sequence of odd numbers

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This can be understood with the difference operator $\Delta f(x) = f(x+1)-f(x)$. Notice each time you apply it the degree drops so you would only be interested in keeping track of the highest degree term, $\Delta(x\mapsto x^n)(x) = nx^{n-1}+\mathcal O(x^{n-2})$. Apply this $n$ times and you'll see what happens. –  Myself Mar 18 '11 at 9:06
See math.stackexchange.com/questions/201330/… for the proof –  Alex Oct 20 '12 at 7:34