Difference of odd squares

Yazan from Dubai International Academy in the United Arab Emirates wrote:

What I found out about subtracting odd squares, is that no matter what odd square numbers you subtract, the answer is always even. For example, let’s take 11 squared take away 9 squared. We know that 11 squared is 121 and 9 squared is 81, so our problem is now:

121 $-$ 81= 40

It is the same with all the other subtractions. All the answers are even.

Rohan from Wilson's School in the UK investigated further (click on the image to see a larger version):

I tested various outcomes for the difference of squares of odd numbers. I noticed that the outcomes are all divisible by 4.

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Sunhari from British School Muscat in Oman used algebra to prove this:

Let the odd numbers be $2n+1$ and $2k+1$ where $2k+1\gt 2n+1,$

$$\begin{array}{rl}(2k+1{)}^2-(2n+1{)}^2& =(2k+1+2n+1)(2k+1-2n-1)\\ & =[2(k+n+1)][2(k-n)]\\ & =(2a)(2b)\text{ for }a=k+n+1\text{ and }b=k-n\\ & =4ab\\ & =4y\text{ for }y=ab\end{array}$$

Roger from Westfield Academy in the UK, Amelie from Jersey College for Girls and Mahdi from Mahatma Gandhi International School in India proved that the difference of two odd squares is actually a multiple of 8. This is Amelie's work (click to see a larger version):

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This is Mahdi's proof (click to see a larger version):

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Rohan and Roger said that the difference of two even squares could be a multiple of 8, but wasn't always. This is Rohan's work (click on the image to see a larger version):

I tested various even numbers using the formula $(a + b) \times (a – b)$ and found that there are many

instances where the answer is a multiple of 8.

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Naima from Bexley Grammar School and Sunhari expressed this algebraically. This is Naima's work:

$a^2 - b^2 = (a + b)(a - b)$

If both $a$ and $b$ are even, $a = 2n$ and $b = 2m$ ($n$ and $m$ being integers)

$a^2 - b^2 = (2n - 2m)(2n + 2m)\

a^2- b^2 = 2(n - m)\times2(n + m)\

a^2 - b^2 = 4(n^2 - m^2)$

This means that the difference between squares of even numbers will always be a multiple of $4.$

Mahdi worked out when the difference will also be a multiple of 8 (click to see a larger version):

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If $a$ and $b$ are both even, then $2a$ and $2b$ are both multiples of 4. If $a$ and $b$ are both odd, then $2a$ and $2b$ are both multiples of 2 but not 4. Can you see how this relates to the pattern in Rohan's table?

Rohan investigated what happens if one number is even and the other is odd (click to see a larger version):

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Sunhari, Naima and Mahdi investigated algebraically whether the difference will be odd or even, but not when it is a multiple of 3. This is Sunhari's work:

Let the numbers be $2k+1$ and $2n$

1. When $2k+1\gt2n$

$$\begin{array}{rl}(2k+1{)}^2-(2n{)}^2& =4k^2+4k+1-4n^2\\ & =4(k^2-n^2+k)+1\\ & =4f+1\text{ for }f=k^2-n^2+k\end{array}$$

2. When $2n\gt2k+1$

$$\begin{array}{rl}(2n{)}^2-(2k+1{)}^2& =4n^2-4k^2-4k-1\\ & =4(n^2-k^2-k)-1\\ & =4g-1\text{ for }g=n^2-k^2-k\end{array}$$

Therefore, an even square subtracted from an odd square gives an odd number which 1 more than a multiple of 4.

An odd square subtracted from an even square gives an odd number which is 1 less than a multiple of 4.

Which numbers can be written as the difference of two squares?

Roger made some interesting observations (click to see a larger version):

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Naima used and generalised this idea:

All odd numbers are the difference of two squares. This is because if $a$ and $b$ are consecutive, the difference between them is 1, and all odd numbers are made up of a consecutive number pair.

 

This is because a consecutive number pair will always contain 1 odd number and 1 even number and so will sum to an odd number. If we express the first number as $n$ and the second as $n + 1,$ their sum will be $2n + 1,$ and so will always be odd. $n$ can be every number, hence the sequence containing every odd number. In other words, the sum of two consecutive numbers will always be part of the $2n + 1$ sequence.

If $a - b = 1$ and $a + b = 2n + 1,$ [then $a^2-b^2$] is simply $2n + 1$; every odd number is possible.

Which numbers can’t we write as the difference of two squares?

Roger wrote:

0, 1 and 2 can't be expressed in this manner. The only negative number which can't be expressed as the difference of 2 squared numbers is -2.

Naima wrote:

We can write all odd numbers and all multiples of 4 as the difference of two squares. However, we cannot write even numbers that are not multiples of 4 as the difference of two squares.

Naima is right. Can you fill in the gaps in this proof to show why?

Imagine that $a^2-b^2 = N.$ We need to show that either $N$ is odd, or $N$ is a multiple of $4.$

Well, if one of $a$ and $b$ is odd and the other is even, then $N$ will be ______.

$a^2-b^2$ can be factorised as _______________.

If $a$ and $b$ are both odd, or both even, then both of the factors of $a^2-b^2$ will be ______.

Therefore $N$ is the product of two ______ numbers, and so $N$ is a multiple of $4.$