What Difference Does It Make?

The Difference Of 2 Squares, that’s what,

A^{2} – B^{2} = O

Yet another hobby project but no programming needed for this; Just simple arithmetic but still interesting.

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I recently ran across the well known “Difference of 2 squares” theorem which is a marvel of simplicity and which states that:

Every positive odd integer can be expressed as the difference of 2 squares. For example:

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

While experimenting with this formula using various combinations of A and B, it looked like

A^{2} – B^{2} = **O**** **would **always** be TRUE if we simply let A = (**O **+ 1) / 2 and B = (**O **– 1) / 2

But the above was just conjecture based on a small sample of an infinite number of possibilities. So the next step was to see if this was “provable.” And yes, it was provable, and the proof is further down below. <=== Click. The proof I arrived at is straightforward and has been around in various forms forever. You can easily find variations of the proof by searching for “Difference of two squares proof.”

That said, the next step, was to see if there was anything to be learned with respect to the **Prime Factors** of the variables A, B, and **O. **While experimenting with various values for A, B, and O, it instantly became clear that sometimes some or all of the Prime Factors of O are not a Prime Factor of A or B. On the other hand, sometimes A, B, and O do indeed have Prime Factors in common. Ultimately, however, **one big thing seemed to stand out** (at least on the surface):

For A^{2} – B^{2} = **O **(i.e. “*the difference of 2 squares*“)…

When A = (**O **+ 1) / 2 and B = (**O – **1) / 2

The Prime Factors of **O **are NEVER one of the Prime Factors of either A or B.

Now we move to proving the above and

investigating its similarity to **Goldbach’s Strong Conjecture**

The above was not expected and had not been proven. However, based on past experience working with the Goldbach Conjecture I was at least keeping my eyes open for it (or some variation of it). Anyway, could this “*Difference of 2 Squares*” situation be similar to what was proved in the post about Goldbach’s Strong Conjecture? In that post we proved that:

For all positive EVEN integers E, the *Prime Factors* of E are * never *one of E’s

*Goldbach Primes …*

An attempt at a “similar” proof, but instead pertaining to “*The Difference Of 2 Squares*” was worth pursuing. At the very least it would be interesting! In other words, what could be said/proven about the Prime Factors of O as it relates to the Prime Factors of A and B? That is, for the “*Difference of 2 Squares*,” can we prove that:

When A = (**O **+ 1) / 2 and B = (**O – **1) / 2

Then the Prime Factors of **O **are NEVER one of the Prime Factors of either A or B?

As luck would have it a proof was not difficult. That proof is down below. <== Click. I have searched to see if anyone else had stumbled into this but have found nothing. On the other hand, I would not be surprised to find out later that others have also been down this road. I’ll keep searching and if you find out something then let me know!

So… the analysis of the ** Goldbach Conjectures **showed that sometimes the variables of a formula NEVER share Prime Factors and sometimes they ALWAYS share at least 1 Prime Factor. And now this analysis of the

**theorem demonstrates something similar (as stated above).**

*Difference Of 2 Squares***HOWEVER**… unlike the

**, the**

*Goldbach Conjectures***theorem doesn’t even mention Prime Numbers!**

*Difference Of 2 Squares*

So, Goldbach’s Strong Conjecture is about EVEN numbers | The Difference of 2 Squares is about ODD numbers |

Goldbach’s is about Prime Numbers
summing to an EVEN number. |
There’s NO MENTION OF Primes
in the Difference of 2 Squares.! |

And yet, in both cases there’s this peculiarity of Prime Factors

of one side of the equation NEVER appearing on the other side.

This whole Prime Number and Prime Factors thing is mysterious! Although we can prove a lot of this stuff the deeper “why” is a mystery. And…, did this whole number/math framework start with the Big Bang? 10 seconds after the Big Bang? Did the “rules” exist before the Big Bang? Were they the same rules?

**What Is Going On!?**

.

Proof — There’s always a solution to “*Difference Of 2 Squares*“

By Using A = (O + 1) / 2 and B = (O – 1) / 2

A^{2} – B^{2} = O and keep in mind for c below that B^{2} = A^{2} – O (after moving the terms around)

- For any O we can let A = (O+1) / 2
- Then A
^{2}= (O^{2}+ 2O + 1) / 4 - Then B
^{2}= A^{2}– O = [ (O^{2}+ 2O + 1) / 4 ] – O - B
^{2}= [ (O^{2}+ 2O + 1) / 4 ] – [4O / 4] - B
^{2}= (O^{2}+ 2O + 1 – 4O) / 4 - B
^{2}= (O^{2}– 2O + 1) / 4 - Now take the square root of the above and we get
- B = (O – 1) / 2

So when A = (O+1) / 2 then B = (O – 1) / 2

Just to make it clear let’s show the complete picture COLOR coded

A^{2} – B^{2} = O

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

Now compute the squares of A and B

[(O^{2} + 2O + 1)/4] – [ (O^{2} – 2O + 1) / 4] = O

Let’s get rid of the 4 on the left side

(O^{2} + 2O + 1) – (O^{2} – 2O + 1) = 4 x O

Start simplifying (removing the parens)

O^{2} + 2O + 1 – (O^{2} – 2O + 1) = 4 x O

O^{2} + 2O + 1 – O^{2} + 2O – 1 = 4 x O

2O + 2O = 4 x O

O = O

.

For the “Difference Of 2 Squares,” A^{2} – B^{2} = O, We Prove That

When A = (**O **+ 1) / 2 and B = (**O – **1) / 2

then the Prime Factors of **O **are NEVER one of the Prime Factors of either A or B.

In the following, the vertical bar means “divides.” E.g. a|b means “a divides b.” This proof is very simple using * Proof By Contradiction*. Let’s begin…

a. O = P_{1} x P_{2}… x P_{n} The “Ps” are the Prime Factors of O

b. A = (O + 1) / 2

c. Let’s assume O and A share a Prime Factor P_{x} . Then P_{x} | A and also P_{x} | O

d. If P_{x} | A then P_{x} also will divide n * A

e. So P_{x} | 2A n=2

f. So P_{x} | 2( (O+1)/2 ) remember that A = (O + 1) / 2

g. So P_{x} | O+1

h. But we also know that** if P _{x} | O then P_{x} can NOT divide O + 1**

So we have a contradiction and we now know that

O and A can NOT share a Prime Factor as stated in c. when A = (O + 1) /2

You can do essentially the same proof for O and B when B = (O – 1) / 2

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