Finite p. divisors

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Source: Test

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#1 h Mar 7, 2004, 4:04 AM • 2 Y

Y by Adventure10, Mango247

Let a,b,c be natural numbers (non zero) ,such that a <sup>n</sup> + b <sup>n</sup> + c <sup>n</sup> has finite prime divisors ,n=1,2....
Prove that :a=b=c

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#2 h Mar 13, 2004, 9:35 PM • 2 Y

Y by Adventure10, Mango247

Let's assume they aren't equal. Then let d=(a,b,c). We can obviously work with a'=a/d, b'=b/d, c'=c/d, which are not all equal to 1, and (a',b',c')=1. Let me call a'=a, b'=b, c'=c (for simplicity).

Let p<sub>i</sub> be the primes wich are the divisors of the numbers of the form S<sub>n</sub>=\sum a<sup>n</sup>, with i from 1 to k. Let n<sub>0</sub>=2\pi (p<sub>i</sub>-1). If one of the numbers a, b, c is divisible by p<sub>i</sub> then S<sub>n<sub>0</sub></sub>=2 (mod p<sub>i</sub>), and if none of the numbers is divisible by p<sub>i</sub>, then S<sub>n<sub>0</sub></sub>=3 (mod p<sub>i</sub>). This means that S<sub>kn<sub>0</sub></sub> is of the form 2<sup>x</sup>*3<sup>y</sup> for all k\in N*. It's easy to see that it's 2 (mod 4) because of the 2 in front of the product, so it has either form 2*3<sup>x</sup> or 3<sup>x</sup>.

I'll finish it a bit later, cause my mom wants some time on the computer, so she's rushing me

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#3 h Mar 14, 2004, 2:54 AM • 2 Y

Y by Adventure10, Mango247

I'll pick up from where I left:

Let a<sup>n<sub>0</sub></sup>=A. In the same way we get B and C from b and c respectively. We have A<sup>k</sup>+B<sup>k</sup>+C<sup>k</sup>=a power of 3 for all k or 2*(a power of 3) for all k. Let's assume it's the first one (the second case is basicly the same).

At the same time, we have (A,B,C)=1. This means that at least 2 of A, B and C are coprime with A<sup>t</sup>+B<sup>t</sup>+C<sup>t</sup>, with t chosen s.t. the sum >3 (such a t exists, because we assume A, B or C>1), so let's assume those 2 are A and B. Take k to be phi(A<sup>t</sup>+B<sup>t</sup>+C<sup>t</sup>). If 3|C we get A<sup>k</sup>+B<sup>k</sup>+C<sup>k</sup>=2 (mod 3<sup>something</sup>), which is false, and if 3 doesn't divide C, we get A<sup>k</sup>+B<sup>k</sup>+C<sup>k</sup>=3 (mod 3<sup>something>1</sup>), so we have a contradiction again.

I think this ends it.

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#4 h Mar 14, 2004, 7:41 AM • 2 Y

Y by Adventure10, Mango247

Dear grobber,
sorry , may be i am stupid but i do not see why Skn_0 has the form 2^x*3^y
Plz explain me,cause really this problem take me so long time though the case of two numbers is easy to see!(This is called Reutter theorem)

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#5 h Mar 14, 2004, 7:54 AM • 2 Y

Y by Adventure10, Mango247

If, for example, a is divisible by p<sub>i</sub>, then b and c aren't (we assume (a,b,c)=1), so b<sup>kn<sub>0</sub></sup>=c<sup>kn<sub>0</sub></sup>=1 (mod p<sub>i</sub>) and c=0 (mod p<sub>i</sub>), so S<sub>kn<sub>0</sub></sub>=2 (mod p<sub>i</sub>).

If, on the other hand, none of a, b or c are divisible by p<sub>i</sub> (this is for a certain i from 1 to k), then a<sup>kn<sub>0</sub></sup>=b<sup>kn<sub>0</sub></sup>=c<sup>kn<sub>0</sub></sup>=1 (mod p<sub>i</sub>), so S<sub>kn<sub>0</sub></sub>=3 (mod p<sub>i</sub>).

This means that the remainder of S<sub>kn<sub>0</sub></sub> (mod p<sub>i</sub>) is either 2 or 3, for all i from 1 to k. If there is a prime divisor of S<sub>kn<sub>0</sub></sub> other than 2 or 3, then it can't be among p<sub>i</sub> with i from 1 to k, so the prime divisors of S<sub>kn<sub>0</sub></sub> aren't contained in the finite set (this was the initial assumption: that all S<sub>t</sub> have their prime divisors in the finite set {p<sub>i</sub> | i from 1 to k}. This shows that the form of S<sub>kn<sub>0</sub></sub> is 2<sup>x</sup>3<sup>y</sup>.

Hope it's correct and clear.

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#6 h Mar 15, 2004, 12:26 PM • 2 Y

Y by Adventure10, Mango247

Well done grobber,i think it's correct!

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