2 unrelated problems

Sunday 12 December 2004 at 2:26 pm | In Articles | 2 Comments
  1. Using the Gelfond-Schneider theorem show that if e is transcendental then \pi is transcendental
  2. If a,b are positive integers such that a>b and a+b is even, show that a^2-a-b^2 is not a square

Notes

  1. This theorem was being discussed at a seminar many years ago as was the transcendence of e. When we were told that it was an obvious consequence that \pi is transcendental we thought that the speaker was joking. When he showed how it followed, there was a moment’s silence and then applause – one of those moments when one realises that mathematics is such a wonderful subject.
    You can find an elementary proof (not using Gelfond-Schneider) of the transcendence of e and \pi here
    It’s interesting to note that e^{\pi} and \pi^e are very close. In fact e^\pi-\pi^e \approx 0.68. See here
  2. I came across this problem when looking through some old papers. I have no idea where it comes from (an old Mathematical Olympiad problem maybe??) or what the solution was.

2 Comments »

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  1. Here’s a solution.
    Suppose that x^2 = a^2 – a – b.
    Then one finds that
    4x^2 + 1 = (2a+2b-1)(2a-2b-1)
    We are assuming that a & b have the
    same parity, and that a>b are positive.
    Thus 2a+2b-1 and 2a-2b-1 are both positive
    integers of the form -1 mod 4. It’s easy to
    see that such numbers have a prime factor
    p of the form -1 mod 4. Thus 4x^2+1 has
    a prime factor of the form p = -1 mod 4.
    Now we are done, since all prime factors of
    4x^2+1 are of the form 1 mod 4.
    (This follows for example by quadratic
    reciprocity).

    Comment by Frank — Friday 18 February 2005 7:44 am #

  2. Wonderful, thank you! Now why didn’t I see that 4x2 + 1 = (2a+2b-1)(2a-2b-1)?

    Comment by Steve — Friday 18 February 2005 2:21 pm #

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