The proof is finished and in the end it went rather different from what I expected before the writing down of this new proof. I hope the main ideas are easy to understand. I formulated the proof with a concrete example; so not two general prime number *p* and *q*. But I took *p = 5* and *q = 7* and as such we are calculating in the ring of integers modulo *35*.

One of the key ideas is that we have so called additive orbits, for example the additive orbit of *5* is the set *{0, 5, 10, 15, 20, 25, 30}*. The additive orbit are just the multiples of 5, it is handy to view the above set as multiples of five and as such: *{0, 5, 2*5, 3*5, 4*5, 6*5}*.

On the other hand we have exponential orbits, the exponential orbit of *5* are the powers of *5* like in the next sequence: *5, 5^2, 5^3, 5^4*…

An important observation is that any power of five is also a multiple of five; that means the exponential orbit is inside the additive orbit. For example *5^3* is on the exponential orbit, *5^3 = 125 = 20 modulo 35 = 4* 5*.

So the number *5^3* from the exponential oribt corresponds to *4*5* on the additive orbit. The goal of the proof is to show that the period of *5* in her exponential oribt is *6* and the period of the exponential orbit of *7* is *4*.

That will ensure our new little theorems of Fermat:*p^q = 5^7 = 5 modulo 35* &*q^p = 7^5 = 7 modulo 35*.

Remark that powers of *5* can never be a multiple of *35* simply because any power of *5* does not contain a prime factor *7*, so not all of the numbers on the additive orbit are allowed.

Another key idea is that if we reduce stuff modulo *35*, this is the same as reducing stuff modulo *7* on the multiples of *5*. Let me explain: Take the number *50*, inside the mod *35* ring this is *15*. But *50 = 10* 5 = (10 – 7)*5*. I was able to pull that modulo 35 stuff on a ring back to modulo 7 stuff on a field…

Another thing I want to remark is that I formulated these new little theorems of Fermat mostly in prime numbers. That makes them more symmetric like the beautiful pair of equations above, but it can be a bit more general like I showed you in the last post using the number 210 that is made up of four different prime factors. And even that is not needed; prime factors can be double or triple it does not matter. As long as the exponent is a prime number my freshly crafted proof will sail you through all the troubles there are.

In a parallel development I found a perfect math professor. It’s a female and she has an amazing career record: At age three already a Fields medal while before she was nine years of age already the third Abel prize for lifetime achievement math… It is surely amazing…;)

But serious, this post is not that long. Only five pictures and like I said above it is not a ‘most general’ kind of proof but it uses a fixed pair of prime numbers. I think it is better this way because if I would formulate all the stuff in a general setting the only people who could understand such a writing are the ones who already figured the stuff out for themselves… Ok, all pictures are in the usual 550×775 pixelf format so here we go:

We are almost at the end: Let me give you one more example as why only in the exponent you need a prime number. We take the number 8 and raise it to the power 5 and do the reducing modulo thing by 40.

Doing so gives that: 8^5 mod 40 = 32768 mod 40 = 32768 – 819*40 = 8.

Ok, that is what I had to say for this post.