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(another)interesting number theory problem

by Mathguy15
Tags: anotherinteresting, number, theory
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Mathguy15
#1
Jan15-12, 04:26 PM
P: 63
a and b are real numbers such that the sequence{c}n=1--->{infinity} defined by c_n=a^n-b^n contains only integers. Prove that a and b are integers.

Sincerely,
Mathguy
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checkitagain
#2
Jan16-12, 04:04 AM
P: 99
Quote Quote by Mathguy15 View Post
a and b are real numbers such that the sequence{c}n=1--->{infinity}
defined by c_n=a^n-b^n contains only integers. Prove that a and b are integers.

Sincerely,
Mathguy
[itex]c_n \ = \ a^n - b^n[/itex]



What about any real numbers a and b, such that a = b, so that [itex]c_n = 0 ?[/itex]
Here, and b don't have to be integers.


Do I have your problem understood, and/or

are there more restrictions on a and b?
Norwegian
#3
Jan18-12, 01:45 AM
P: 144
I assume you mean a≠b.

Since a-b and a2-b2=(a-b)(a+b) are both integers, a+b is rational, and we get a and b are rational.

We can write b=m/t and a=(m+kt)/t with (m,t)=1. Assume t≠1, then there is an integer s such that k is divisible by ts but not by ts+1.

Let p be a prime larger than t and 2s+2.

cp=ap-bp=(pktmp-1+k2t2(...))/tp

Both the second term and the denominator are divisible by t2s+2, while the first term is not, so the fraction is not an integer. It follows that t=1 and we are done.

dodo
#4
Jan18-12, 06:43 AM
P: 688
(another)interesting number theory problem

Quote Quote by Norwegian View Post
a+b is rational, and we get a and b are rational.
Sorry, Norwegian, but why? For example, sqrt(2) and 3-sqrt(2) are both irrational, and they add up to 3.
micromass
#5
Jan18-12, 07:32 AM
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P: 18,086
Quote Quote by Dodo View Post
Sorry, Norwegian, but why? For example, sqrt(2) and 3-sqrt(2) are both irrational, and they add up to 3.
Both a+b and a-b are rational. So (a+b)+(a-b)=2a is rational.
dodo
#6
Jan18-12, 11:18 AM
P: 688
Ahhh, thanks, Micromass.
dodo
#7
Jan19-12, 03:48 AM
P: 688
By the way, this is a beautiful proof, and I'm still trying to figure out how did you come to it, Norwegian.

I presume you started from both ends. At the finishing end, you needed a^n-b^n to be a rational but not an integer. At the starting end, the way you expressed a=b+k suggests the use of the binomial theorem to evaluate powers of b+k (or powers of the numerator of it). If a and b are rational, then a^n and b^n (with a=b+k) were going to end up having a common denominator, so you concentrated in making the numerator of a^n-b^n a non-integer. Then divisibility / factorization issues enter; though I still don't see in which order did (1) finding the largest power of t dividing k, (2) coprimality conditions, and (3) finding a prime p that does not divide most of the things around, in which order these three came to be, and what suggested them.

I always find instructive to see the genesis of proofs; it adds to the inventory of ways of constructing new ones.
Mathguy15
#8
Jan23-12, 04:56 PM
P: 63
Quote Quote by checkitagain View Post
[itex]c_n \ = \ a^n - b^n[/itex]



What about any real numbers a and b, such that a = b, so that [itex]c_n = 0 ?[/itex]
Here, and b don't have to be integers.


Do I have your problem understood, and/or

are there more restrictions on a and b?
Oh sorry! Yes, a and b had to be distinct.
Mathguy15
#9
Jan23-12, 04:58 PM
P: 63
Quote Quote by Norwegian View Post
I assume you mean a≠b.

Since a-b and a2-b2=(a-b)(a+b) are both integers, a+b is rational, and we get a and b are rational.

We can write b=m/t and a=(m+kt)/t with (m,t)=1. Assume t≠1, then there is an integer s such that k is divisible by ts but not by ts+1.

Let p be a prime larger than t and 2s+2.

cp=ap-bp=(pktmp-1+k2t2(...))/tp

Both the second term and the denominator are divisible by t2s+2, while the first term is not, so the fraction is not an integer. It follows that t=1 and we are done.
That is an interesting proof, and I will take the time to digest it later! Thanks


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