**“Even or Odd?”**

What difference does it make?

“Half the difference.”

About a month ago I wrote about “* The Difference of 2 Squares*” where it’s discussed how every positive ODD integer can be expressed as the difference of 2 squares. For example:

O = A^{2 }– B^{2 } (where “Oh” is any Odd positive integer).

You can read that article <here>.

That said, about a week ago I pondered whether a similar statement could be made about **EVEN** numbers**.** For example, a “*difference of 2 squares*” or something closely related except for EVEN numbers.

What I found is surprising. Interested? If yes then read on…

**Lock The Gates!**

So, while experimenting with various numbers and “formulas” it quickly surfaced that **EVERY** positive **EVEN** integer **E** can * ALWAYS* be expressed as the Difference of 2 Squares divided by 2. For example:

A^{2} – B^{2}

E = * ————* This is VERY similar to the Difference of 2 Squares for ODD integers which, again is: O = A

^{2 }– B

^{2 }

2

Also based on experiments with 20 or so cases it seemed that it is ALWAYS possible to select an appropriate A and B that makes the above formula true just by using this simple formula:

**A = (E/2) + 1**

**B = (E/2) – 1**

**Note:** It **also** seemed to be true that, except for 2, the Prime Factors of E are **NEVER** one of the Prime Factors of either A or B! **More on that later.**

To summarize and clarify, in this project it was conjectured by way of experiment that:

Every positive **EVEN** integer **E** can * ALWAYS* be expressed as the

*. For example:*

**Difference of 2 Squares divided by 2** A^{2} – B^{2}

E = ————

2

And that we can ALWAYS accomplish that by choosing an appropriate A and B according to the formula:

A = (E/2) + 1

B = (E/2) – 1

Of course the above was just conjecture. What was needed was a proof. And since a proof was relatively easy for *The Difference of 2 Squares* for ODD numbers it might be equally easy for this similar (yet different) case for EVEN numbers.

Anyway, the proof of the above is down below. Click <**here**> to jump to it Proof1.

Now for a different but related topic…

Prime Factors of

E, A, and B

A^{2} – B^{2}

For E = ———— Experimentation indicated that:

2

Except for 2, ALL of the Prime Factors of **E** are **NEVER** one of the Prime Factors of either **A** or **B**!

This appears to be the same situation as for the *Difference Of 2 Squares* for ODD integers. I.e., ALL of the Prime Factors of O are likewise never a Prime Factor of either A or B (which was __ proven__ for the

*Difference of 2 Squares*for ODD numbers).

But for EVEN numbers this was still just conjecture based on 20 or so experimental values for E, A, and B. But it might be possible to prove it. And indeed, it was possible to prove that, except for 2, EVERY Prime Factor of E is NEVER a Prime Factor of either A or B. The proof is below. Click <here> to jump to it. proof 3

**Now for still another discovery!**

Consider the following:

We knew that we can express every positive Odd integer as **O** = **A**^{2 }**– B**^{2 }** **

Also, we know (it is obvious) that each EVEN positive integer E is the **previous O** + 1 as well as the **succeeding O** – 1. Like so…

**E = O + 1 and E = O – 1**

And since each O is the *difference of 2 squares *we have:

**E = A ^{2} – B^{2} + 1 and E = A^{2} – B^{2} – 1**

So in some sense, we can express all EVEN numbers as the * difference of 2 squares *for adjacent ODD numbers

**+-**1 (as shown above). But this is kind of trivial. It’s like saying every EVEN number is some adjacent ODD number

**+-**1. However, what

**IS**interesting for

__of these cases (plus 1 and minus 1) is that:__

**BOTH**

**Each Prime Factor of E is ALWAYS a Prime Factor of A**

**and yet,**

**Each Prime Factor of E is NEVER a Prime Factor of B**

Did I emphasize that this applies to BOTH cases?

This is easily proved and you can jump to the proof by clicking <here> Proof 2

**Epilogue**

I’ve spent a couple of hours trying to find whether others have stumbled on this “* Difference of 2 Squares Divided by 2” for EVEN integers* but have not found anything yet. I’m surprised by that but nonetheless I strongly suspect others must have been down this path. Anyway, if you find (or know of) something then please let me know.

It’s interesting how the relationships of variables in formulas often will ALWAYS or NEVER share Prime Factors (as opposed to SOMETIMES). As shown here, we can PROVE these things but at a deeper level the * WHY *is still a mystery!

**Open The Gates!**

**History**

**2 Photos of the original WhiteBoard notes for this project follow.**

**================= Proofs Follow ==================**

.

.

**Proof 1**

**Every positive EVEN integer E **

**can be expressed as **

**the Difference of 2 Squares divided by 2**

**like so…**

**E = (A ^{2} – B^{2}) / 2**

As it turns out, **this is always true when we let**

**A = (E/2) + 1**

**B = (E/2) – 1**

It’s also **sometimes** true for some other settings of A and B for some values of E. For an example, just pick **any** pair of EVEN positive integers letting A be the greater and B being the lesser. For example:

Let A=20 and B=5.

Then E = (400 – 25)/2 = 192

Anyway, and once again…

**When we choose**

**A = (E/2) + 1**

**B = (E/2) – 1**

Then using those values we can ALWAYS create

any **E = (A**^{2}** – B**^{2}**) / 2**

To prove all of this we will use ** proof by contradiction**. We start by assuming that this is FALSE. That is, we assume that for any E, and using the settings for A and B as described above, then we will end up with

**E** **!=** **(A**^{2}** – B**^{2}**) / 2**

Let’s see if this is * even* possible (a pun!).

Let’s compute the right hand side of the equation.

We’ll start by squaring A

A^{2} = ( (E/2) + 1) x ( (E/2) + 1)

A^{2 }= E^{2}/4 + 2E/2 + 1

Now we’ll square B.

B^{2} = E^{2}/4 – 2E/2 + 1

Now compute A^{2} – B^{2}

A^{2}–B^{2} = (E^{2}/4 + 2E/2 + 1) – (E^{2}/4 – 2E/2 + 1)

A^{2}–B^{2} = 4E/2 = 2E

Now we finally finish computing the right hand side of the equation…

(A^{2}-B^{2})/2 = 2E/2 = E

So… when we choose A and B as described above we always get E on both sides of the equation. Which contradicts the assumption of

E **!=** (A^{2} – B^{2}) / 2

You can also help convince yourself by trying a few different values for E (and hence for A and B also).

————— End of Proof ———–

.

.

**Proof 2**

**For**

**E = A**^{2}** – B**^{2}** +- 1 **

**or in other words**

**E = O +- 1**

**Each Prime Factor of E is ALWAYS a Prime Factor of A**

**and yet,**

**Each Prime Factor of E is NEVER a Prime Factor of B**

Let’s choose the case of E = O + 1 (leaving E = O – 1 for later). Then…

**O = ****A**^{2}** – ****B**^{2 }** (**the * Difference of 2 squares* for ODD integers.)

E = O + 1

then

**E = ****A**^{2}** – ****B**^{2}** + 1 **** (****the diff of 2 squares for O, then + 1**** )**

Also remember from the article about * Difference Of 2 Squares* for

**odd**integers that we can choose A and B according to the formulas:

A = (O+1)/2

B = (O-1)/2

Using these values for A and B we can then say

E = ( (O+1)/2 )^{2} – ( (O-1)/2 )^{2} + 1

But also realize that O+1 = E and so O–1 = E-2

so… when we evaluate the above and we get

E = (E/2)^{2} – ( (E-2)/2 )^{2} + 1

E = (E/2)^{2} – ( (E/2 – 1 )^{2} + 1

At this point we are able to see that except for 2, each Prime Factor of E will indeed divide E/2 which is A. This means that

**Except for 2, ALL Prime Factors of E**

**are ALSO a Prime Factor of A.**

And e can also now see that none of the Prime Factors of E can divide (E/2) – 1 which is B. This means that

**Except for 2, NONE of the Prime Factors of E**

**are a Prime Factor of B.**

If we go thru the same steps we can similarly prove the same thing for the case

E = O – 1 = **A**^{2}** – B**^{2}** – 1 **

but I won’t do it here.

**————- End of Proof ———–**

.

.

**Proof 3**

For

**E = (A ^{2} – B^{2}) / 2**

**when **

**A = (E/2) + 1**

**B = (E/2) – 1**

**Then**

**Except for 2,**

**EVERY Prime Factor of E is **

**NEVER a Prime Factor of either A or B.**

Using just the Prime Factors of E …

E = 2^{n} x P_{1}^{n} x P_{2}^{n }x P_{3}^{n }x … P_{z}^{n}

and

A = (E/2) + 1

then

A = 2^{n} (P_{1}^{n} x P_{2}^{n }x P_{3}^{n }x … P_{z}^{n })

————————————— + 1

2

where P_{1}..P_{z} are positive ODD integers and exponent n is any positive integer though it may be different for any/all of the Prime Factors 2 and P1..Pz. This leads to:

A = 2^{n-1} x (P_{1}^{n} x P_{2}^{n }x P_{3}^{n }x … P_{z}^{n }) + 1

So, if we now divide the above by any of the Prime Factors of E (e.g. P_{3}) then we’ll get the following:

.

2^{n-1} x (P_{1}^{n} x P_{2}^{n }x P_{3}^{n }x … P_{x}^{n }) 1

————————————— + ———

P_{3} P_{3}

.

The left side of the + sign is an integer.

The right side of the + sign is NOT an integer.

This tells us that P_{3} (__which is a Prime Factor of E__) is NOT also a Prime Factor of A since it can NOT evenly divide A (because of the 1/P_{3}).

And we’ll get the same result for all of the Prime Factors of E except (sometimes) for 2.

—————– End Of Proof 3 —————

———————– End of Article ———————–