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.

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have you tried graphing it

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

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.

>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.

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.

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

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

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).

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

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.

n = 6ab + a + b

n = a(1 + 3b) + b(1+3a)

Obviously there are infinitely many (a,b) that aren't solutions here.

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

Hint modulo 3

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

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

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)

Interesting, thank you

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

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.

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'll look into that

>it has a lot to do with the twin prime conjecture

It literally is just the twin prime conjecture, only phrased more annoyingly.

>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

and

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?

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.

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

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.