Two S-hypotheses (Sorokin’s hypothesis)

[Victor Sorokine]

I advance two similar S-hypotheses (“Sorokin’s hypothesis”).

 

S-hypothesis-1 (confirmed by concrete calculations on the computer):

In the equality [math]A^{n^t}+B^{n^t}=(A^{n^{t-1}}+B^{n^{t-1}})R[/math], where

- prime n> 2,

- integers A and B are relatively prime,

(t-1)-digit end of each prime divider of the number R (excluding n) in the base n is equal to 1.

For example (for n=3): [math]8^3-1=(8-1)*73=(8-1)*(8*9+1) [/math]; [math]8^3+1=(8+1)*57=(8+1)*(3*19)=(8+1)*[3*(2*9+1][/math]; and so forth.

 

S-hypothesis-2 (more questionable):

In the equality [math]A^n+B^n=(A+B)R[/math], where

- prime n> 2,

- integers A and B are relatively prime,

- their k-digit ends are the k- digit ends of some numbers of a^n and b^n,

- the k-digit end of the number R is equal to 1,

the k-digit end of each prime divider of the number R (excluding n) in the base n is equal to 1.

[Victor Sorokine]

Theorem about the dividers (Sorokin’s theorem)

 

If integers A, B and A-B are relatively prime and not multiple prime n>2, and the t-digits end of the number R in the equality [math]A^n-B^n=(A-B)R[/math] in the base n is equal to 1,

then the t-digits end of of each prime divider m of the number R is also equal to 1.

 

Proof consists of several theorems.

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Merged post follows:

Consecutive posts merged

I approach the publication of strict proofs of theorems.

 

Theorem 1.

 

If integers A, B and A-B are relatively prime and are not multiple by prime n>2, then the last digit of each prime divider m of the number R in the equality [math]A^n-B^n=(A-B)R[/math] is equal to 1 in the base n.

 

Proof.

 

It is known that under theorem conditions the number R is not divided by n and numbers A-B and R have no common divider.

 

Let us show now that also the prime number q of the form of [math]q=p+1[/math], where p is not divided by n, is the divider of the number [math]A^n-B^n[/math] and is not the divider of the number R.

 

According to Fermat's little theorem, number [math]A^p-B^p[/math] is divided by q.

 

It is utilized the solution of the following linear diophantine equation: nx-py=1[/math].

 

The number [math]A^ {nx}-B^ {nx}[/math] is divided by m. But [math]A^{nx}-B^{nx}=A^{py+1}-B^{py+1}=(A^p)^y-(B^p)^y[/math], where the numbers [math]A^p[/math] and [math]B^p[/math] have (according to Fermat's little theorem) last digit equal to 1. Consequently, the number A-B is divided, but the number R is not divided by q.

 

Thus, the numbers of form [math]q=p+1[/math], where p is not divided by n, are not the dividers of the number R. QED.