
#1
Jan1512, 04:26 PM

P: 63

a and b are real numbers such that the sequence{c}n=1>{infinity} defined by c_n=a^nb^n contains only integers. Prove that a and b are integers.
Sincerely, Mathguy 



#2
Jan1612, 04:04 AM

P: 99

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? 



#3
Jan1812, 01:45 AM

P: 144

I assume you mean a≠b.
Since ab and a^{2}b^{2}=(ab)(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 t^{s} but not by t^{s+1}. Let p be a prime larger than t and 2s+2. c_{p}=a^{p}b^{p}=(pktm^{p1}+k^{2}t^{2}(...))/t^{p} Both the second term and the denominator are divisible by t^{2s+2}, while the first term is not, so the fraction is not an integer. It follows that t=1 and we are done. 



#4
Jan1812, 06:43 AM

P: 688

(another)interesting number theory problem 



#5
Jan1812, 07:32 AM

Mentor
P: 16,545





#6
Jan1812, 11:18 AM

P: 688

Ahhh, thanks, Micromass.




#7
Jan1912, 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^nb^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^nb^n a noninteger. 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. 



#8
Jan2312, 04:56 PM

P: 63





#9
Jan2312, 04:58 PM

P: 63




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