Some not formally proven discovery

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Some not formally proven discovery

The number $(a+1)^{n}) - a^{n}$ (where a is any positive integer, and n > 1 is prime number), if not being prime,
has as its divisor(s)/factor(s) the member(s) of arithmetic progression with step 2n.
n = 3 -> {7,13,19,25,31,37,43,49,55,61,67,73,…}
n = 5 -> {11,21,31,41,51,61,71,81,91,101,…}
n = 7 -> {15,29,43,57,71,85,99,113,127…}
n = 11 -> {23,45,67,89,111,133,155…}
and so on …..

Programmatically (I used a JavaScript code) this discovery was proved for different range of numbers a and n.
Mblacob

Posts: 6
Joined: Thu Jun 14, 2018 6:58 am
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Re: Some not formally proven discovery

Mblacob wrote:The number $(a+1)^{n} - a^{n}$ (where a is any positive integer, and n > 1 is prime number), if not being prime,
has as its divisor(s)/factor(s) the member(s) of arithmetic progression with step 2n.
n = 3 -> {7,13,19,25,31,37,43,49,55,61,67,73,…}
n = 5 -> {11,21,31,41,51,61,71,81,91,101,…}
n = 7 -> {15,29,43,57,71,85,99,113,127…}
n = 11 -> {23,45,67,89,111,133,155…}
and so on …..

Programmatically (I used a JavaScript code) this discovery was proved for different range of numbers a and n.

Mblacob

Posts: 6
Joined: Thu Jun 14, 2018 6:58 am
Reputation: 1

Re: Some not formally proven discovery

This can be proven to be true (except for $n=2$ where it is false).

If all prime divisors of $(a+1)^p -a^p$ are of the form $2pk+1$, then it implies all factors are of this form (because factors are just a product of prime divisors and because $(2pk_1+1)(2pk_2+1) = (2pk_3+1)$ where $k_3 = 2pk_1k_2+k_1+k_2$).

It is easy to check that 2 is not a divisor of $(a+1)^p -a^p$, so any prime divisor $q$ must be odd. If $q$ is of the form $pk+1$, then it must also be of the form $2pk'+1$ (since $p$ and $q$ are odd).

So it is enough to show that a prime $q$ that divides $(a+1)^p -a^p$ must be of the form $pk+1$ (for $p$ an odd prime).

Suppose $q$ is a prime that divides $(a+1)^p -a^p$ (but we don't yet know it's form).
This implies $(a+1)^p = a^p \bmod{q}$, which means $(a+1)^{pm} = a^{pm} \bmod{q}$ for any choice of positive integer $m$.

Fermat's little theorem tells us that $b^{r(q-1)+1} = b \bmod{q}$ for any integer $b$, prime $q$, and positive integer $r$.

So if we could choose $m$ such that $pm$ is of the form $r(q-1)+1$, then by Fermat's little theorem, we'd get
$(a+1)^{pm} = a^{pm} \bmod{q}$ implies $a+1 = a \bmod{q}$ which is clearly false. Consequently we know that $pm$ can never be of the form $r(q-1)+1$, or put another way there is no $m$, such that $pm=1 \bmod{(q-1)}$.

If there exists such an $m$ it would be known as the multiplicative inverse of $p$ under mod $(q-1)$. It is a well known theorem that the multiplicative inverse exists if and only if the highest common factor of $p$ and $q-1$ is $1$. Since we know the inverse doesn't exist and the only factors of $p$ are $1$ and $p$ (because it is prime) we must have the highest common factor is $p$.

Finally, this means that $p$ is a factor of $q-1$, therefore $q-1=kp$ for some $k$, proving $q$ is of the form $pk+1$.
Guest

Re: Some not formally proven discovery

I have to thanks the Guest, who formally proved my statement (or theorem, if it possible to say).
Any way, it's allowed to create Prime Numbers Generator. Generator (for given a and n) builds the sequence of prime numbers, and some of them are Primary Prime and others are Secondary Prime, as results of sequential division of not prime number by knowing factors for given n.
Some examples of Primary Prime numbers generated are: 1517045588059, 2257404775627, 3142080149209, 5311018201441, 6391813317473, 2019169299698041.
I have a programming code to support calculation, but it is not optimized by running time yet.

Mblacob

Posts: 6
Joined: Thu Jun 14, 2018 6:58 am
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