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?
I assume you mean a≠b. Since a-b and a^{2}-b^{2}=(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 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^{p-1}+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.
Sorry, Norwegian, but why? For example, sqrt(2) and 3-sqrt(2) are both irrational, and they add up to 3.
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.