# Proof Regarding Property of Odd Integers [duplicate]

The question I am working on is:

"Use a direct proof to show that every odd integer is the difference of two squares."

Proof:

Let n be an odd integer: $n = 2k + 1$, where $k \in Z$

Let the difference of two different squares be, $a^2-b^2$, where $a,b \in Z$.

Hence, $n=2k+1=a^2-b^2$...

As you can see, this a dead-end. Appealing to the answer key, I found that they let the difference of two different squares be, $(k+1)^2-k^2$. I understand their use of $k$--$k$ is one number, and $k+1$ is a different number--;however, why did they choose to add $1$? Why couldn't we have added $2$?

-

## marked as duplicate by Amzoti, Henry T. Horton, Fabian, Micah, Davide GiraudoFeb 13 '13 at 17:09

Adding the 1 is the correct way to start... try it out. Simplify (k+1)^2-k^2 and see if you get something familiar. – kaine Feb 13 '13 at 14:42
@kaine No, I understand that is very helpful to use 1, but why couldn't we have chosen 2? – Mack Feb 13 '13 at 14:43
if you use 2 it becomes 4k+4 which is not an odd number; if you use one (k+1)^2-k^2 = 2k+1. You can extract something interesting from the 2 but it won't be the odd number proof you are looking for. – kaine Feb 13 '13 at 14:45
"why did they choose to add 1?" Because it was something simple to try. And when they tried it, it worked! – Hurkyl Feb 13 '13 at 15:13
I'm not sure this is a duplicate – I think the questioner is misunderstanding what they are being asked, rather than failing to find an answer. See my answer, which would be completely inappropriate for the other question. – Ben Millwood Feb 14 '13 at 23:38

What's going on here, I think, is you're confused about what you're being asked to prove.

The statement is "every odd integer is the difference of two squares", or, more precisely, "for all odd integers $n$, there exist $a$ and $b$ such that $a^2 - b^2 = n$". Think for a bit about what the difference between "for all..." and "there exists..." is, and maybe you'll realise what's going on.

You have to come up with a proof that works for every odd integer, so you start with an arbitrary choice $n = 2k + 1$. But then you seem to go on to pick an arbitrary difference of two squares, $a^2 - b^2$. But that's not necessary – we're not trying to prove anything about every difference of two squares, just one, so you only need to show that there is a difference of two squares that satisfies the condition, and in particular you choose it, it isn't arbitrary.

Think of the statement as "if you give me an odd integer $n$, I can give you $a$ and $b$ such that $a^2 - b^2 = n$". In particular, you can give me whatever odd integer you like, but then I get to choose what $a$ and $b$ are. In particular, I can choose $a$ to be $b + 1$ if I like.

We have to show that $a^2 - b^2$ is an odd number

But in general that's just not true. For example, $4^2 - 2^2 = 12$. So if you've found yourself trying to prove that statement, you've clearly gone wrong somewhere.

-
Thank you. That was quite helpful. – Mack Feb 15 '13 at 11:42

Another approach, a graphical proof: $$\underbrace{\begin{array}{ccccc} 1\odot & 3\otimes & 5\odot & \cdots & (2k - 1)\otimes\\\hline \bigodot & \bigotimes & \bigodot & \cdots & \bigotimes\\ \bigotimes & \bigotimes & \bigodot & \cdots & \bigotimes\\ \bigodot & \bigodot & \bigodot & \cdots & \bigotimes\\ \vdots & \vdots & \vdots & \ddots & \vdots\\[1ex] \bigotimes & \bigotimes & \bigotimes & \cdots & \bigotimes\\ \end{array}}_{\textstyle k}$$

From this we have $k^2 = 1 + 3 + 5 + \dotsb + (2k - 1)$. Hence, \begin{align} 2k + 1 &= \Big(1 + 3 + 5 + \dotsb + (2k - 1) + (2k + 1)\Big) - \Big(1 + 3 + 5 + \dotsb + (2k - 1)\Big)\\ &= (k + 1)^2 - k^2 \end{align}

-
+1: Re-purposing the picture-proof for "The sum of consecutive odd integers, starting at 1, is a square." Nice! – Blue Feb 13 '13 at 14:55
I believe a proof of this sort is a little beyond my abilities, at this point in time. – Mack Feb 13 '13 at 14:57
I have edited the answer to explain how the $(k + 1)^2 - k^2$ thing arises from this. Does that help? – Herng Yi Feb 13 '13 at 15:00
A classic, from Fibonacci's The Book of Squares (Liber Quadratorum) – Math Gems Feb 13 '13 at 18:47

Directly:

$$2k+1=k^2+2k+1-k^2=(k+1)^2-k^2\ldots$$

-

Here you have another approach...

Note that $$\sum_{i=1}^n { i } = \frac{n(n+1)}2$$ so that the sum of the first $n$ odd naturals is $$\sum_{i=1}^n { (2i-1) } = 2\left( \sum_{i=1}^n { i } \right) - n = n(n+1)-n = n^2$$ We have showed that the sum of the first $n$ odd naturals is $n^2$. The $n$-th odd natural is, trivially, equal to the sum of the first $n$ minus the sum of the first $n-1$: $$2n-1 = \sum_{i=1}^n { (2i-1) } - \sum_{i=1}^{n-1} { (2i-1) }$$ and from what we found, we can write it as $$2n-1 = n^2 - (n-1)^2$$

-
+1 Ah, I loved this one... – DonAntonio Feb 13 '13 at 15:17
A classic, from Fibonacci's The Book of Squares (Liber Quadratorum) – Math Gems Feb 13 '13 at 18:49
@MathGems Wow, didn't know that. Thanks for pointing it out! – AndreasT Feb 13 '13 at 23:15

Note that $a^2 - b^2 = (a + b)(a - b)$. Solve the simultaneous equations $a + b = n$ and $a - b = 1$.

This is where you got $(k + 1)^2 - k^2$ from - $(a + b)(a - b)$ then matches to $((k + 1) + k)((k + 1) - k)$.

-
What are you setting $a^2 - b^2$ to get the simultaneous equations $a+b=n$ and $a-b=1$? – Mack Feb 13 '13 at 14:51
I notice that I have a lot of freedom in choosing what $(a + b)$ and $(a - b)$ will be, so I purposely give them nice values to ensure that the product will give $n$ while $a$ and $b$ are positive. The obvious way to do it is to let the larger one $a + b = n$ and the smaller one $a - b = 1$. – Herng Yi Feb 13 '13 at 14:56
How is it that you have freedom in that choice? We have to show that $a^2-b^2$ is an odd number; so, we assume that that is true. Since $n$ is some arbitrary odd number ($n=2k+1$), then we set $a^2-b^2$ to equal to $n$: $a^2-b^2 = 2k+1$ – Mack Feb 13 '13 at 14:59
Sure you could do that, but my approach is also valid because $a$ and $b$ are two variables so it's fine to restrict them with two equations. In fact, the $(k + 1)^2 - k^2$ thing probably came from observing the results of solving $a + b = n$ and $a - b = 1$. – Herng Yi Feb 13 '13 at 15:04

$$2k + 1 = 1 \times (2k +1) = (\color{#C00}{k + 1} -{\bf k}) (\color{#C00}{k + 1} +\bf{ k})$$

-