A little problem

Posted on 31/05/2021 by Peter Cameron

In connection with the power graphs of unitary groups, I came across the following little number-theoretic conundrum. Can anyone solve it?

Let q be an odd power of 2 (bigger than 2). Show that (q2−q+1)/3 is not a prime power unless it is a prime.

About Peter Cameron

I count all the things that need to be counted.
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6 Responses to A little problem

  1. Ofir Gorodetsky says:
    31/05/2021 at 10:09

    I do not know how to solve the problem. However, if we write q=2^k for odd k, I can show that if (q^2-q+1)/3 is a prime power then either k is a prime or k is a power of 3.

    I am going to use Bang’s Theorem, which is a special case of the better known Zsigmondy’s theorem. It is an elementary theorem with useful consequences.
    It says that a number of the form 2^n-1 has a prime divisor p which does not divide any of the numbers 2^1-1, 2^2-1, …, 2^(n-1)-1, except for the case n=6.

    Let us write q as 2^k for odd k. The expression (q^2-q+1) is 2^(6k)-1, divided by (2^(3k)-1)(2^(2k)-1)(2^2-1)/(2^k-1).
    For any number n such that n divides 6k but does not divide 3k, 2k, our number is divisible by 2^n-1.
    (A useful way to verify this is to use the factorization 2^n-1 = prod_{d | n} phi_d(2) where phi_d is the d’th cyclotomic polynomial. Bang equivalently says there is a prime dividing phi_n(2) which does not divide phi_d(2) for d<n, unless n=6.)

    Now let s be a divisor of k, such that s is coprime to 3 and not equal to k itself.
    Then n=6k/s divides 6k but not 3k or 2k (since s is odd and coprime to 3). Note also that 6k/s is not equal to 6. Hence there is a prime number p_n dividing your expression, coming from the choice n=6k/s, at least under our assumptions.

    So if your expression is not a prime power, necessarily k has at most one divisor which is not equal to k itself and is coprime to 3. One such divisor is 1. So there can't be other such divisor, which means k is a power of 3, or a prime.

    Reply
  2. Ofir Gorodetsky says:
    31/05/2021 at 10:33

    As for the original problem: It seems to be accessible using deep methods from number theory.
    E.g., Bugeaud, Mignotte and Siksek (Annals, 2006) proved that the only perfect powers in the Fibonacci sequence are 0, 1, 4, 8, 144, and their methods are related to those underlying the solution of Fermat’s Last Theorem.
    Their problem is closely related to yours: both the Fibonacci sequence and the sequence a_k = (2^(2k)-2^k+1)/(2^2-1) are example of Lucas sequences of the first kind.
    To solve your problem it suffices to determine the prime powers in the sequence a_k (we may restrict to a_(2k+1)), which might be even more accessible than the Fibonacci case, but I know very little about this particular topic.

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    • Ofir Gorodetsky says:
      31/05/2021 at 12:18

      Apologies, I wasn’t accurate when saying that a_k = (2^(2k)-2^k+1)/(2^2-1) is a Lucas sequence. However, I still believe these kind of methods should be useful in this exponential Diophantine equation. There are also some more accessible papers by Bugeaud, Mignotte and others on these topics (e.g. Bugeaud, Mignotte and Roy, ‘On the Diophantine equation (x^n – 1)/(x-1) = y^q’, Pacific Journal of Mathematics, 2000).

      Reply
  3. Nick Gill says:
    10/11/2021 at 11:06

    Nagell has a paper on this precise equation. The paper is “Des equations indetermines x^2+x+ 1=y^n et x ^2+x+1=3y^n”. Norsk matematisk forenings skrifter. Serie I Nr. 2 (I921).

    Nagell’s paper is cited by Ljunggren in his paper “Einige Bemerkungen…”

    I can’t access Nagell’s paper so can’t tell you if it gives the answer you want!

    Reply
  4. volkan says:
    13/01/2022 at 12:35

    any progress on these numbers?

    thanks

    Reply

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