Divisor of the form an+1
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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.
a>0 and n>1
I still have hard time to explain it.
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?
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?
Re: Divisor of the form an+1
Here are few examples :
3^72^7=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)
3^72^7=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)
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.
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.
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
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 ab=n
4^31=641=63=7*9 (7=3*2+1)
Thank you Notzeb.
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 ab=n
4^31=641=63=7*9 (7=3*2+1)
Thank you Notzeb.
Re: Divisor of the form an+1
@notzeb
The theorem 4 is about a^n1 not a^n+b^n or a^nb^n
It is well known and the proof was yet given here.
The theorem 4 is about a^n1 not a^n+b^n or a^nb^n
It is well known and the proof was yet given here.
Re: Divisor of the form an+1
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 q1, 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:The theorem 4 is about a^n1 not a^n+b^n or a^nb^n
I'm not sure how to parse this sentence.Goahead52 wrote:It is well known and the proof was yet given here.
Zµ«VjÕ«ZµjÖZµ«VµjÕZµkVZÕ«VµjÖZµ«VjÕ«ZµjÖZÕ«VµjÕZµkVZÕ«VµjÖZµ«VjÕ«ZµjÖZÕ«VµjÕZµkVZÕ«ZµjÖZµ«VjÕ«ZµjÖZÕ«VµjÕZ
Re: Divisor of the form an+1
Thank you.
I understand now why.
I can go forward.
I understand now why.
I can go forward.
Re: Divisor of the form an+1
I want your expertise.
We know the factorization of :
a^3+b^3= (a+b)(a^2ab+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
We know the factorization of :
a^3+b^3= (a+b)(a^2ab+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

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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^2ab+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.
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^2ab+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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