## Divisor of the form an+1

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Goahead52
Posts: 431
Joined: Thu Oct 16, 2014 9:28 am UTC

### Divisor of the form an+1

There is at least one prime divisor of the form an+1 dividing (3^n)-(2^n)
a>0 and n>1
I still have hard time to explain it.

Goahead52
Posts: 431
Joined: Thu Oct 16, 2014 9:28 am UTC

### Re: Divisor of the form an+1

It works with 3^n+2^n too.
Will it work with a^n+b^n (gcd(a,b)=1)?
If it is the case then it will explain the Last Fermat Theorem with elementary tools.
What do you think about that?

Goahead52
Posts: 431
Joined: Thu Oct 16, 2014 9:28 am UTC

### Re: Divisor of the form an+1

Here are few examples :

3^7-2^7=29*7 (29=7*4+1 and 71=7*10+1)
3^7+2^7=5*463 (463=7*66+1)
5^10+3^10=2*17*101*2861 (101=10*10+1 and 2861=256*10+1)
8^11+5^11=13*23*28892183 (23=11*2+1 and 28892183=11*2626562+1)

Goahead52
Posts: 431
Joined: Thu Oct 16, 2014 9:28 am UTC

### Re: Divisor of the form an+1

Either my claim is wrong then I need a counterexample to find out why
Either my claim is true and maybe there is an elementary proof to that or no one has a clue.
I sent few examples to build my case. There are many others.
Any comment will be welcomed.

notzeb
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### Re: Divisor of the form an+1

Check out this short proof of Zsigmondy's theorem, especially Theorem 4. I think it should answer your question.
Zµ«V­jÕ«ZµjÖ­Zµ«VµjÕ­ZµkV­ZÕ«VµjÖ­Zµ«V­jÕ«ZµjÖ­ZÕ«VµjÕ­ZµkV­ZÕ«VµjÖ­Zµ«V­jÕ«ZµjÖ­ZÕ«VµjÕ­ZµkV­ZÕ«ZµjÖ­Zµ«V­jÕ«ZµjÖ­ZÕ«VµjÕ­Z

Goahead52
Posts: 431
Joined: Thu Oct 16, 2014 9:28 am UTC

### Re: Divisor of the form an+1

I have found counterexample :

2^3+1^3 = 9=3*3 (not divided by 3a+1)

here a+b=n
2+1=3
It explain maybe the counterexample.
But is works with a-b=n

4^3-1=64-1=63=7*9 (7=3*2+1)

Thank you Notzeb.

Goahead52
Posts: 431
Joined: Thu Oct 16, 2014 9:28 am UTC

### Re: Divisor of the form an+1

@notzeb

The theorem 4 is about a^n-1 not a^n+b^n or a^n-b^n
It is well known and the proof was yet given here.

notzeb
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### Re: Divisor of the form an+1

Goahead52 wrote:The theorem 4 is about a^n-1 not a^n+b^n or a^n-b^n
The same argument works for a^n - b^n as long as a and b are relatively prime (the argument is fairly easy so that article left it out: just consider the order of a/b in the multiplicative group (mod q), note that this order must divide q-1, and that the order is n if q divides the nth cyclotomic polynomial evaluated at a/b but does not divide n). For the case of a^n + b^n, you should meditate on the problem yourself.
Goahead52 wrote:It is well known and the proof was yet given here.
I'm not sure how to parse this sentence.
Zµ«V­jÕ«ZµjÖ­Zµ«VµjÕ­ZµkV­ZÕ«VµjÖ­Zµ«V­jÕ«ZµjÖ­ZÕ«VµjÕ­ZµkV­ZÕ«VµjÖ­Zµ«V­jÕ«ZµjÖ­ZÕ«VµjÕ­ZµkV­ZÕ«ZµjÖ­Zµ«V­jÕ«ZµjÖ­ZÕ«VµjÕ­Z

Goahead52
Posts: 431
Joined: Thu Oct 16, 2014 9:28 am UTC

### Re: Divisor of the form an+1

Thank you.
I understand now why.
I can go forward.

Goahead52
Posts: 431
Joined: Thu Oct 16, 2014 9:28 am UTC

### Re: Divisor of the form an+1

I want your expertise.

We know the factorization of :

a^3+b^3= (a+b)(a^2-ab+b^2)

We know that :

a^3+b^3 is divisible by a prime number of the form 3x+1

Hence we could show that :
the equation a^3+b^3=c^3 can not hold

So the Fermat Last theorem with n=3 could be proved using elementary tools.
We could do the same with a^n+b^n=c^n where n>3

PsiSquared
Posts: 126
Joined: Wed May 09, 2012 6:02 pm UTC

### Re: Divisor of the form an+1

Goahead52 wrote:I want your expertise.

We know the factorization of :

a^3+b^3= (a+b)(a^2-ab+b^2)

We know that :

a^3+b^3 is divisible by a prime number of the form 3x+1

Hence we could show that :
the equation a^3+b^3=c^3 can not hold

How? I don't really see any connection.

Goahead52
Posts: 431
Joined: Thu Oct 16, 2014 9:28 am UTC

### Re: Divisor of the form an+1

PsiSquared wrote:
Goahead52 wrote:I want your expertise.

We know the factorization of :

a^3+b^3= (a+b)(a^2-ab+b^2)

We know that :

a^3+b^3 is divisible by a prime number of the form 3x+1

Hence we could show that :
the equation a^3+b^3=c^3 can not hold

How? I don't really see any connection.

I need some long calculations to show the link.
I will post later my approach at least for n=3
Do you agree that a prime number of the form 3x+1 always divides a^3+b^3 (gcd(a,b)=1)?

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