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(another)interesting number theory problem

  1. Jan 15, 2012 #1
    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.

  2. jcsd
  3. Jan 16, 2012 #2
    [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?
  4. Jan 18, 2012 #3
    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.


    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.
  5. Jan 18, 2012 #4
    Sorry, Norwegian, but why? For example, sqrt(2) and 3-sqrt(2) are both irrational, and they add up to 3.
  6. Jan 18, 2012 #5
    Both a+b and a-b are rational. So (a+b)+(a-b)=2a is rational.
  7. Jan 18, 2012 #6
    Ahhh, thanks, Micromass.
  8. Jan 19, 2012 #7
    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.
    Last edited: Jan 19, 2012
  9. Jan 23, 2012 #8
    Oh sorry! Yes, a and b had to be distinct.
  10. Jan 23, 2012 #9
    That is an interesting proof, and I will take the time to digest it later! Thanks
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