6

Consider this equation where $a$ and $b$ are positive integers.

$$k = \frac{2^a - 1}{2^{a+b} - 3^b}$$

This equation has the trivial solution $k=1, a=1, b=1$.

How would I find more solutions, or show that no more exist? I'm not asking anyone to solve it, but just explain how I might explore the solutions myself.

I came up with this equation as one whose solution would imply one specific kind of cycle in the Collatz iteration. Since the Collatz conjecture is believed to hold, I expect that there is proof that this equation has no other solutions, and am interested to see what mathematical techniques can be used to eliminate just this one case.

Gottfried Helms
  • 34,920
CSStudent7782
  • 185
  • Yes, but if you're asking people to solve an equation whose solution would only yield one of infinitely many possible counter-examples, and that equation is certainly not a new idea, don't you think all the experts already tried it? If so, and you're still asking here, it is really off-topic. Anyway, in case you are curious to read more, you can start with this paper. – user21820 Jul 31 '21 at 12:51
  • You're welcome! If you've any other mathematical curiosities, feel free to ask me in chat. – user21820 Jul 31 '21 at 13:22
  • Don't want to say much, but I like it that/how another one came up to a valid, and interesting, formula for discussing the problem (loops-in-collatz). Mind-of-a-pioneer, even if it is only at the first steps. – Gottfried Helms Jul 31 '21 at 19:57
  • @user21820 do you agree with the reason for closure? "This question is not about mathematics, within the scope defined in the help center." I read said scope, and it's hard to understand how my question isn't about mathematics. – CSStudent7782 Aug 04 '21 at 13:12
  • Concerning the closure of this question, it is because attempts at solving open problems are considered off-topic on Math SE, even though on the surface it is about mathematics. If you wish to confirm this, you can ask a moderator in the Mods' Office. – user21820 Aug 04 '21 at 13:43
  • 1
    @user21820 in that case, it may have been closed in error. Perhaps I asked it poorly, but I was really hoping someone would answer the question related to divisibility. Although do go deeply into where that equation comes from (an open problem), what I really wanted to know is how we might find out about integers solutions to an equation like that. – CSStudent7782 Aug 04 '21 at 15:19
  • 1
    @CSStudent7782: Ah I see. Then I know what you can do. You should put your question (the equation you are interested in) as your main question (at the top). And then you put your motivation below, explaining that you knew this will not solve Collatz but you are interested in the number theoretic techniques that we can use to exclude this case. I think if you do that it will be readily reopened. – user21820 Aug 04 '21 at 16:10
  • The reason it is bad in its current form is that the title is "This equation describes a subset of the possible counter-examples to the Collatz Conjecture, can non-trivial Integer solutions be found", which directly implies that you want people to find counter-examples to the Collatz conjecture via the equation you found. So please fix your title as well to reflect what you (now) want to ask about. After you have fixed your question with regard to these, feel free to ping me here or in chat and I will add a reopen-vote. – user21820 Aug 04 '21 at 16:12
  • 3
    @lonestudent - yes,I've arrived at the same formula, only with different names for the involved variables. And I've tried to derive from this a counter-argument against the existence of loops. But no approach sufficed, except that argument from transcendence theory which low-bounds the distance between powers of 3 and powers of 2. That argument has first been developed by Ray Steiner in 1976 against the existence of 1-cycles. Simons/deWeger made this a bit better readable later and also found better low-bounds than Steiner had available. (...) – Gottfried Helms Aug 04 '21 at 18:58
  • 3
    (...) See my workout of this at http://go.helms-net.de/math/collatz/Collatz_1cycledisproof.pdf on my webspace (My old collatz-article is at http://go.helms-net.de/math/collatz/Collatz061102.pdf but today I'd say, it is awful to read...) – Gottfried Helms Aug 04 '21 at 18:59
  • Btw. did you notice, that it has another solution $k=-1$ for $(a,b)=(1,2)$? – Gottfried Helms Oct 02 '21 at 22:55

2 Answers2

4

There is another representation of this formula, which connects it with an -as well unsolved- "detail in the Waring-problem" (see mathworld.com) namely the distance of $(3/2)^N$ to the next integers.

By rewriting (I use different variables names from your formula to match the notation in an essay of mine:N for b - because "N" indicates the "N"umber of odd steps, "A" for "a" because I use capital letters for exponents in this environment, see section 4.2 in the essay) $$ k(2^A2^N-3^N)=2^A-1 \\ 2^A(2^N k-1)=3^N k-1 \\ $$ $$ 2^A={3^N k-1 \over 2^N k-1} \tag 1$$ Introducing a functional notation $$ f(N,k) = {3^N k-1 \over 2^N k-1} \tag 2 $$ $\qquad \qquad $ relates to a well-known conjecture, but which is again unproven until today.
$\qquad \qquad $ We can reformulate this as $$ f(N,k) = (3/2)^N + (3/4)^N/k - \varepsilon_{N,k} \tag 3 $$ $\qquad \qquad $ where $\varepsilon_{N,k} < 1/2^N$ for $N,k \gt 1$.

Looking at the first part only and omitting the small subtractive summand, we find an expression, which occurs in the "detail" $$ f^*(N,k) = (3/2)^N + (3/4)^N \lt \lceil (3/2)^N \rceil \qquad \text {for } N \gt 7 \tag 4 $$ and that means, that in $(1)$ not only we cannot have a perfect power of $2$ on the lhs, but even not an integer at all, because due to conjecture $(4)$ the next integer above $(3/2)^N$ is larger than $f^*(N,k)$ and thus than $f(N,k)$ as well.

So if the "detail in the Waring-problem" could be solved/proved independently, then again we had the argument against the "1-cycle".

I didn't investigate the reverse idea: but I think it should be an interesting discussion, whether the Steiner/Simons/deWeger-disproof of the "1-cycle" can be expanded into a formal solution of the "detail in the Waring-problem". Perhaps this is doable with more experience/math-training than I have.

Appendix: see the picture of $f^*(N,1)$ . (The picture is rescaled for the $\tanh^{-1}()$ of the interval $0 .. 1$ of the resp. fractional parts)

Indeed, for $N \gt 7$ the red points for $f^*(N,1)$ (denoted as $g(N)$ in the picture) lay in the very near of the grey/blue points, and for $N \gt 50$ the different coordinate is practically indiscernable, thus empirically confirming the conjecture $(4)$ from the "detail" up to $N=20000$. picture

Gottfried Helms
  • 34,920
2

Looking at old entries in my literature-database, I found an interesting limiting formula for a lower bound of $2^{a+b}-3^b$. Some short tinkering with it seem to show, that you can prove your conjecture for all $a+b > 27$ with it.
The formula is (Stroeker/Tijdeman,'71): $$ \mid 2^x - 3^y \mid \gt \exp(x (\log 2- \frac1{10})) \qquad \text{for all } x,y \in \mathbb N \quad \text{and } x\gt27 \quad \;^{[1]}\tag 1$$ This can be applied to your equation. By (1) we can write $$ 2^{a+b}-3^b \gt \mu ^{a+b} \qquad \text{where } \mu =1.80967483607... \tag 2$$ and thus $$ k = { 2^a-1\over 2^{a+b}-3^b} \lt { 2^a\over \mu^{a+b}} \qquad \text{for }a+b\gt 27 \tag 3$$ Here the rhs can be found to be smaller than $1$ for $(a+b) \gt 27$: $$ \text{(rhs)}=\exp( a\cdot \ln2 - (a+b)(\ln2 - 1/10))\\ =\exp(0.1 a-( \ln2-0.1)b) \\ \approx \frac{1.1^a}{1.8^b} \lt 1 \tag 4$$ and thus we can conclude $$\implies k \lt 1\qquad \text{for }a+b\gt 27 \tag 5$$

$\qquad\qquad\quad$(Hope I didn't mess with signs and computation...)

Someone might even brush this up a bit...

$\;^{[1]}$The citation of formula (1) is from

R.J.STROEKER & R.TIJDEMAN 
Diophantine equations (with appendix by P.L.Cijsouw, A.Korlaar & R.Tijdeman) 
in: MATHEMATICAL CENTRE TRACTS 154, COMPUTATIONAL METHODS IN NUMBER THEORY; PART I;
MATHEMATISCH CENTRUM, AMSTERDAM  1982

and they attribute this result to W.J.Ellison in 1970/1971

[25] ELLISON,W.J., Recipes for solving diophantine problems by Baker's method,
Sèm.Th.Nombr.,1970-1971,Exp.No.11, Lab.Thèorie Nombres,
C.N.R.S.,Talence,1971.
Gottfried Helms
  • 34,920