A Difficult Math Problem

Hello anons of IQfy, there is this math question that I've been trying to answer for the longest time and I wanted to see if any of you were smart enough to solve it.

The problem in question is the following: "Are there infinite positive integers that are not the solution to 'abs(6ab)+a+b' being a and b two integers different from 0?"

For example, 13 is abs(6*2*-1)+2-1 but 7 cannot be obtained from that formula. To solve this problem, you have to either prove that there are infinite numbers like 7 or that there aren't.

I have been trying to solve this problem on and off for a few years, and I want to see if there's someone here capable of doing it. If you give me a proof, you will have my deepest gratitude.

Thalidomide Vintage Ad Shirt $22.14

Tip Your Landlord Shirt $21.68

Thalidomide Vintage Ad Shirt $22.14

  1. 4 weeks ago
    Anonymous

    have you tried graphing it

    • 4 weeks ago
      Mathanon

      I have. I've used graphic calculators and I've also placed the numbers in a grid. I could see some patterns but none of those allowed me to solve the problem. Additionally, the problem with graphing is that it works best against problems that also use rational numbers, but here we only care about integers.

    • 4 weeks ago
      Mathanon

      https://i.imgur.com/4Tl8zRP.png

      I have. I've used graphic calculators and I've also placed the numbers in a grid. I could see some patterns but none of those allowed me to solve the problem. Additionally, the problem with graphing is that it works best against problems that also use rational numbers, but here we only care about integers.

      And here you have the grid

      • 4 weeks ago
        Anonymous

        The way you worded your question is confusing. Do you think this is an accurate wording of the conjecture?

        Conjecture: There exists an infinite number of positive integers that are not the solution to the following equation: f(a,b) = abs(6ab) + a + b, where a and b are integers not equal to 0.

        • 4 weeks ago
          Mathanon

          >Conjecture: There exists an infinite number of positive integers that are not the solution to the following equation: f(a,b) = abs(6ab) + a + b, where a and b are integers not equal to 0.
          Yes, that is what I meant.

          >Do you think this is an accurate wording of the conjecture?
          I tried to explain it as clearly as I could. I thought it was accurate enough, so sorry for doing a poor job at that.

      • 4 weeks ago
        Anonymous

        You only need to prove that when growing that window it will contain numbers above the amount of numbers that can be placed in that window, so there will be always numbers missing in that window: those who satisfy the inequality. Of course the annoying part is to formalize it.

        • 4 weeks ago
          Mathanon

          Yes, there will always be numbers missing. For example, there are 400 numbers in the window of that image, around 200 if you remove duplicates. 200<620, so yes, there are some missing numbers.
          But that doesn't prove it. I've tried that same thing many times before.

          >so there will be always numbers missing in that window: those who satisfy the inequality
          There are some numbers missing, for example 109, 148 or 301, but that doesn't mean that they satisfy the inequality.
          109 = 6*22*1 -22 -1
          148 = 6*21 +21 +1
          301 = 6*60 -60 +1

          • 4 weeks ago
            Mathanon

            I forgot the times 1. Here you go

            109 = 6*22*1 -22 -1
            148 = 6*21*1 +21 +1
            301 = 6*60*1 -60 +1

      • 4 weeks ago
        Anonymous

        I tried a lattice point counting method but it is inconclusive.
        Basically, let f(a,b) = 6|ab|+a+b.
        The goal is to count all of the (a,b) such that f(a,b) <= N and prove this is much less than N.
        We can bound f below by 4|ab| so counting the (a,b) such that 4|ab|<=N will give an over estimate.
        We just need to do this for one quadrant then multiply by 4.
        Using https://en.wikipedia.org/wiki/Divisor_summatory_function
        the rough total is 4*D(N/4).
        This is inconclusive since D(x) ~ x*log(x).

        • 4 weeks ago
          Anonymous

          m8 it's just a troll version of the twin prime conjecture and you're not going to solve it with shit from wikipedia

  2. 4 weeks ago
    Mathanon

    [...]

    If I understand this correctly, you are saying that there aren't enough different numbers in a square of side 2k to invalidate all numbers below a certain value. While that is true, there may be some numbers outside that square that invalidate them.
    Take for example k=10.
    If we take the 3k^2 different numbers inside that square, que get 300, but in the corner we have a 620, so there must be some numbers that cannot be described by the formula as you said.
    However, I think you didn't take into account that if min(|a|,|b|) < k < max(|a|,|b|) they can also contribute to the different number count. a=12 b=2 returns 146, for example.
    Correct me if I misunderstood what you said, but I don't think that proof is correct.
    Nontheless, I appreciate that you took the time to think about it a bit.

  3. 4 weeks ago
    Anonymous

    n = 6ab + a + b
    n = a(1 + 3b) + b(1+3a)
    Obviously there are infinitely many (a,b) that aren't solutions here.

    • 4 weeks ago
      Mathanon

      I'm sorry, but I fail to see how that's obvious.
      Also, you didn't take into account the absolute value, so a(1+3b)+b(1+3a) will only return the values that can be found on the upper right of

      https://i.imgur.com/56xDQ52.png

      [...]
      And here you have the grid

      • 4 weeks ago
        Anonymous

        Hint modulo 3

  4. 4 weeks ago
    Anonymous

    I ran it up to n=ten million, and the density is approaching 0.26 * n^0.86, there oughta be a proof

    • 4 weeks ago
      Anonymous

      how do you run it that high? How can you test for all possible values of a and b?

      • 4 weeks ago
        Anonymous

        first - instead of running for negative a, and b, just loop over positive and check four values, --, -+, +-,++
        So for a>=1 b>=1:
        n in {6ab-a-b, 6ab-a+b, 6ab+a-b, 6ab+a+b}

        Note that frexample a=7, b=4 has the same results as a=4,b=7, so you only have to test one - arbitrarily pick a as >= b

        Then loop over
        a 1..10_000_000
        b 1..a

        Then note that the lowest n value for any (a,b) is 6ab-a-b
        so now
        max_n=10_000_000
        for a 1..max_n
        ..for b 1..a
        .... if 6ab-a-b > max_n: break
        ......count[6ab-a-b]++
        ......count[6ab-a+b]++
        ......count[6ab+a-b]++
        ......count[6ab+a+b]++

        And the lowest n for any a is when b=1, so 5a-1, so you only need to run up to max_n/5+1
        for a 1..(max_n/5+1)

    • 4 weeks ago
      Mathanon

      Interesting, thank you

      Hint modulo 3

      Ok, I'll think about it some more time.

      https://i.imgur.com/A4Ue8Sv.png

      Instead of using absolute value assume a,b > 0 and separate into 3 cases

      6ab+a+b = 1/6 * ((6a+1)(6b+1)-1)
      6ab-a-b = 1/6 * ((6a-1)(6b-1)-1)
      6ab-a+b = 1/6 * ((6a+1)(6b-1)+1)

      or

      6n+1 = (6a+1)(6b+1)
      6n+1 = (6a-1)(6b-1)
      6n-1 = (6a+1)(6b-1)

      Iff either 6n+1 or 6n-1 is composite one of these equations will give a,b > 0 from a nontrivial factorization. I.e. n is representable iff 6n-1 and 6n+1 aren't twin primes.

      And that is precisely what I'm working on, the twin prime conjecture. I don't have very high hopes of solving such a famous problem, but I'm having fun trying.

      m8 it's just a troll version of the twin prime conjecture and you're not going to solve it with shit from wikipedia

      Yes, it has a lot to do with the twin prime conjecture, but I am no troll. I just wanted to see if there was some easy solution that I was missing and that maybe someone here could find. 6|ab|+a+b looks like something that should have a very simple answer, but as I've said, I haven't found one in the last few years.

      I tried a lattice point counting method but it is inconclusive.
      Basically, let f(a,b) = 6|ab|+a+b.
      The goal is to count all of the (a,b) such that f(a,b) <= N and prove this is much less than N.
      We can bound f below by 4|ab| so counting the (a,b) such that 4|ab|<=N will give an over estimate.
      We just need to do this for one quadrant then multiply by 4.
      Using https://en.wikipedia.org/wiki/Divisor_summatory_function
      the rough total is 4*D(N/4).
      This is inconclusive since D(x) ~ x*log(x).

      I'll look into that

      • 4 weeks ago
        Anonymous

        >it has a lot to do with the twin prime conjecture
        It literally is just the twin prime conjecture, only phrased more annoyingly.

        • 4 weeks ago
          Mathanon

          >It literally is just the twin prime conjecture, only phrased more annoyingly
          If I had just said "Solve the twin prime conjecture for me" I wouldn't have received any interesting responses, but by phrasing it that way, which is the furthest I have gotten in that problem, I have received help from people like

          I ran it up to n=ten million, and the density is approaching 0.26 * n^0.86, there oughta be a proof

          and

          I tried a lattice point counting method but it is inconclusive.
          Basically, let f(a,b) = 6|ab|+a+b.
          The goal is to count all of the (a,b) such that f(a,b) <= N and prove this is much less than N.
          We can bound f below by 4|ab| so counting the (a,b) such that 4|ab|<=N will give an over estimate.
          We just need to do this for one quadrant then multiply by 4.
          Using https://en.wikipedia.org/wiki/Divisor_summatory_function
          the rough total is 4*D(N/4).
          This is inconclusive since D(x) ~ x*log(x).

          I wanted to present the problem in a way that would match more closely the way I'm trying to work with it.

          Also, maybe there's someone who wouldn't have been able to solve the conjecture on their own but was able to solve the 6|ab|+a+b, who knows?

  5. 4 weeks ago
    Anonymous

    Instead of using absolute value assume a,b > 0 and separate into 3 cases

    6ab+a+b = 1/6 * ((6a+1)(6b+1)-1)
    6ab-a-b = 1/6 * ((6a-1)(6b-1)-1)
    6ab-a+b = 1/6 * ((6a+1)(6b-1)+1)

    or

    6n+1 = (6a+1)(6b+1)
    6n+1 = (6a-1)(6b-1)
    6n-1 = (6a+1)(6b-1)

    Iff either 6n+1 or 6n-1 is composite one of these equations will give a,b > 0 from a nontrivial factorization. I.e. n is representable iff 6n-1 and 6n+1 aren't twin primes.

    • 4 weeks ago
      Mathanon

      Also, that is a very interesting way of linking it to the twin primes. They way I did it was very different and I think much more complicated that it needed to be. Thank you

    • 4 weeks ago
      Anonymous

      Bump.
      I also assume a,b> 0 with three cases like . I think the problem is similar the the thing that every number can be represented with a cofactor of 6 + 1 or something. I remember it vaguely but it was basically able to construct every number of the positive integers. If you can show an algorithm that you construct every possible intenger with ab +a+b then you have a proof that the inequality is false. I am working on that and will report later.

Your email address will not be published. Required fields are marked *