Divisibility by p^p: A, B, C Sol'ns & More

[Terry Coates]

I have found how to get three integers A B and C such that A^p - B^p - C^p is of form N*p^p with p > 2 and N not divisible by p+2.or p.
This is A = p^(p-1) , B = A-1, C = 1 . This works with p = 3 , 4 and 5.
My questions are: does it work with all values of p > 2 and is there any other way of achieving these properties?

 

[Terry Coates]

Sorry not with p = 4, but probably with all primes.

 

[fresh_42]

For $p=4$ I get $1.024.254$ for the sum which is not divisible by $4^4=256\,$.

 

[fresh_42]

I have found how to get three integers A B and C such that A^p - B^p - C^p is of form N*p^p with p > 2 and N not divisible by p+2.or p.
This is A = p^(p-1) , B = A-1, C = 1 . This works with p = 3 , 4 and 5.
My questions are: does it work with all values of p > 2 and is there any other way of achieving these properties?

You get it for all odd $p$ which can be proven by the expansion $(a-1)^n=\sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k}a^k\,$. You simply subtracted the terms for $k=0$ and $k=n$ and the rest is divisible by $a$ and by $n$.

 

[Terry Coates]

It also works for all values of B and C with B + C = A as long as B and C are not divisible by p. eg for p= 3 B,C = 8,1 or 7,2 or 5,4
Numerical examples for p higher than 3 are not possible to verify, due to the very large values involved. However from an algebraic expression it is easily shown how this is true for all prime values of p

You may have guessed my interest is in Fermat's last theorem, and the findings here seem to be the simple proof of this, since the condition of divisibility by p^p is necessary if A^p - B^p - C^p = 0 , and this divisibility is only possible in the way I show, where the sum is always greater than zero.
So have I hit on Fermat's proof? That is for all prime values of p which is considered all that is needed. (p = 4 having been proved di8fferently)

 

[fresh_42]

It also works for all values of B and C with B + C = A as long as B and C are not divisible by p. eg for p= 3 B,C = 8,1 or 7,2 or 5,4
Numerical examples for p higher than 3 are not possible to verify, due to the very large values involved.

They are, but what for? You may find as many examples as you want, and still have no proof.

However from an algebraic expression it is easily shown how this is true for all prime values of p

When I read sentences like this, I usually immediately stop reading. In nine of ten cases, it is wrong what's hidden behind easily. If it was so easy, why don't tell us?

You may have guessed my interest is in Fermat's last theorem, ...

... which has been proven by Andrew Wiles and history as well as proof are strong indications, that there is no short way to it.

... and the findings here seem to be the simple proof of this, since the condition of divisibility by p^p is necessary if A^p - B^p - C^p = 0 , and this divisibility is only possible in the way I show, where the sum is always greater than zero.

Unproven statement and irrelevant to the case.

So have I hit on Fermat's proof?

No. I dare to claim this without any inspection. One of the few cases in math, where evidence is almost as good as a proof.

That is for all prime values of p which is considered all that is needed. (p = 4 having been proved di8fferently)

Sorry, but the chances to find a simple proof for FLT are literally zero.

 

[mfb]

You have shown that there are specific numbers A,B,C such that A^p-B^p-C^p is a non-zero value that has some known prime factor. That doesn't tell us anything about other A,B,C.