Something I didn't think of.

 

I’m sorry.  There is something I didn’t think of some days ago, while I was looking for a simple solution to Fermat’s statement.

I had just proved (you can click here to see it) that any triangle that has sides a, b, c the length of positive integers satisfies Fermat’s statement (aⁿ ≠ bⁿ + cⁿ for any positive integer value of n > 2).  And I thought that was it…

A mistake.  I mean, my proof was solid and sound; but I hadn’t finished yet.  Why?

All triangles verify that each of its three sides is smaller than the sum (addition) of the other two.

Let’s forget triangles from this point on.

If we have three positive integers a, b, c… 

One of them could be equal to the sum of the other two as in a = (b + c).

One of them could be bigger than the sum of the other two as in a > (b + c).

I need to prove that those two possibilities actually satisfy Fermat’s statement in order to have a complete, simple solution.

And that’s what I’m going to do in this blog post.

CASE 5: Three positive integers a, b, c and a = (b + c).

If a = (b + c), then…

(I lacked some symbols -at the text processor in my laptop- I needed for these part of the proof, so I wrote it on a notebook page and scanned it.  Please do excuse my penmanship):

So aⁿ > bⁿ + cⁿ → aⁿ ≠ bⁿ + cⁿ for any integer value of n > 2.  Which is Fermat’s statement in all its splendor.

Then I tried to prove (which was ridiculous because there was no need to since it is exactly the same case) what would happen if instead of a being equal to (b + c), it was b the one that was equal to (a + c).

Sigh.  A little trap I fell into, courtesy –I guess- of plain fatigue & some deprivation of sleep because of this long heatwave we’re suffering nowadays.

Well, I’ll show you where I got to.

(And again, not enough symbols for my typing and Carlos goes handwriting…)

Right.  Statement satisfied. 

Not a big surprise, huh?

CASE 6.  Three positive integers a, b, c and a > (b + c).

I want you to understand this: a is a positive integer, a number… and  (b + c) is another positive integer, another number.

Only that a is a bigger number than (b + c).

Then…  Think about it.  For a positive integer value of n…  Which is bigger, aⁿ or (b + c)ⁿ?

Come on, this is simple.  I guess you can trust your intuition here…

That’s right, aⁿ is bigger than (b + c)ⁿ.  Or, written in Mathematics language, aⁿ > (b + c)ⁿ.

The trick I’m going to use is…  What’s the result for (b + c)ⁿ?

Look, I’m old and sometimes I’m not sure about formulas I learnt a long, long time ago…  But I’m quite sure that when you solve (b + c)ⁿ, you end up obtaining a bⁿ, a cⁿ and –between the two of them- a lot of other products of positive integers by powers of b and c (combined or by themselves).

Yes, that’s right: aⁿ is bigger than bⁿ plus cⁿ plus a lot of other positive integer products…

So aⁿ > bⁿ + cⁿ for an integer value of n equal or bigger than 1 →

→ aⁿ ≠ bⁿ + cⁿ for an integer value of n equal or bigger than 1 →

→ aⁿ ≠ bⁿ + cⁿ for an integer value of n > 2.

Finally, I think I can go for

Quod erat demonstrandum.

 

Thank you for reading all this.  Have a great summer,

CARLOS.