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Let

$$t = \frac{3^4 + 2^k(3^3 + 2^l(3^2 + 2^m(3^1 + 2^n)))}{2^{p + n + m + l + k} - 3^5}$$

where k, l, m, n, p can only be natural numbers >=1.
The unknown t has also to be a natural number >= 1.

While this has the trivial solution t = 1 for k = l = m = n = p = 2, I am wondering if there is any way to prove the uniqueness of this solution.
Looking at all these prime numbers at different powers, I could try to relate this to a vector space base which should imply that the numerator and the denominator are linearly independent, but since I am no expert in this, I need your help at least to know what direction to follow.

Thank you,
Catalin

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  • $\begingroup$ Is $3^0$ intended ? $\endgroup$ – Yves Daoust Dec 13 '19 at 10:00
  • $\begingroup$ Hi Yves, it is a typo. It should be a 1 to match the pattern. I fixed that. Thanks for the comment. $\endgroup$ – Catalin Neacsu Dec 14 '19 at 16:13
  • $\begingroup$ Looking at it again, I could relate this to an extension of Fermat's theorem too. A crazy idea though. Note that power "p" appears only in the lower term. $\endgroup$ – Catalin Neacsu Dec 14 '19 at 16:17
  • $\begingroup$ Hmm, isn't it just the question, what the numbers $t$ are that are members of an "$5$-odd-step" cycle in the Collatz problem? It can with really elementary means be proved, that such a cycle except for $t=1$ cannot exist in the positive integers. It needs only a little productformula and an enumeration of candidates in small numbers, say $t<100$ or so. (I added also the tag collatz-problem to the tag-list) $\endgroup$ – Gottfried Helms Dec 15 '19 at 23:38
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Because I know your equation from the consideration of cycles in the Collatz-problem (*) I'll put it into my own framework of formulae and extract an answer from that.
Let us write one "odd step" of the Collatz-transformation from positive odd $a$ to positive odd $b$ by $$ b = {3a + 1\over 2^A} \qquad \text{ where $A$ is that $b$ is odd} \tag 1 $$ Then take this to $5$ transformations writing $$ b = {3a + 1\over 2^A} \qquad c = {3b + 1\over 2^B} \qquad d = {3c + 1\over 2^C} \qquad e = {3d + 1\over 2^D} \qquad f = {3 e+ 1\over 2^E} \qquad \tag 2 $$ Then we could as well write the form which notes only the exponents to make this all shorter $$ f=C(a;A,B,C,D,E) \tag 3 $$ and use the parameters of the transformation $N=$number-of-exponents$=$number-of-odd-steps and $S=$sum-of-exponents$=$number-of-oven-steps .
With this my exponents $(A,B,C,D,E)$ agree to your numbers $(k,l,m,n,p) $ and $N$ is your exponent $5$ at the $3$ in the denominator. My $S$ is your sum in the exponent at the $2$ in the denominator.

We can write the trivial product $$ b \cdot c \cdot d \cdot e \cdot f =\left(3a+1 \over 2^A\right) \left(3b+1 \over 2^B\right)\left(3c+1 \over 2^C\right)\left(3d+1 \over 2^D\right)\left(3e+1 \over 2^E\right) \tag 4 $$ and because we assume it's a cycle $a=f$ and we can rewrite and reorganize to the following "critical formula for cycles": $$ a \cdot b \cdot c \cdot d \cdot e =\left(3a+1 \over 2^A\right) \left(3b+1 \over 2^B\right)\left(3c+1 \over 2^C\right)\left(3d+1 \over 2^D\right)\left(3e+1 \over 2^E\right) \tag 5 $$ and then $$ 2^S = \left(3+\frac1a\right)\left(3+\frac1b\right)\left(3+\frac1c\right)\left(3+\frac1d\right)\left(3+\frac1e \right) \tag 6 $$ Because all numbers $1 \le a,b,c...,e \lt \infty$ each parenthese on the rhs lies between $ 3 \lt \left(3+\frac1a\right) \le 4$ thus $$ 3^N \lt \text{rhs} \le 4^N$$ and thus the we must also have $$3^N \lt \text{lhs}=2^S \le 2^{2N} $$ and thus $ 3^5=243 \lt 2^S \le 1024 =2^{10}$ which implies $ 8 \le S \le 10$

The key idea of the following is:

  • Let us assume that $a$ is the smallest element, then it must be smaller than some average number $\alpha = \text{mean}(a,b,c,d,e) = (a+b+c+d+e)/5$

Now solving your problem by enumeration begins.

  • We test $S=8$
    So $2^S = 256$ the next larger perfect power of $2$ larger than $3^N$.
    Let us determine a solution where all numbers are equal and likely noninteger (but not $1$ because this is the trivial cycle with $S=2N=10$) $$ 2^8 = \left(3+\frac1 \alpha \right)^5 \\ \alpha= { 1\over 2^{8/5} -3 } \approx 31.8 \tag 7 $$ Thus $a \le \left\lfloor \alpha \right\rfloor =31$ and we need test manually all of the candidates for $a$. That means $a \in \{5,7,11,13,17,19,23,25,29,31 \} $ . (Note, that $3 \not \mid a$ because no number divisible by $3$ can be result of an odd-step-transformation, and note that we moreover could exclude more numbers automatically from the candidate-set)
    Actually testing all that possible candidates for $a$ shows that
    there is no nontrivial cycle with $N=5$ and $S=8$ .

  • We test $S=9$
    Using $S=9$ and $\text{lhs}=512$ gives an $\alpha=2.07$ which means that $a=1$ is required in that case, but we excluded that case because we do not want the trivial cycle.

  • We need not test $S=10$
    We need not test $S=10$ because that means only the "trivial cycle" $a=b=c=d=e=1$ and $A=B=C=D=E=2$

Putting that all together this proves that

Theorem 1 there is no other solution besides the trivial one for the $5$-odd-step-cycles and thus no nontrivial solution to your formula in the question.

You see this is a fairly general ansatz which can disprove many $N$-odd-step cycles manually by few enumerations, and thus your formula in the question can be evaluated in much more generalized fashion.

Additional remark: Of course we did not need to test this actually, because it is already known that all numbers $a \lt 10^{20}$ (or so) converge to $1$ and thus cannot be member of a nontrivial cycle. But I assume you wanted a proof which you can check manually


(*)(I've also tagged your question accordingly, hoping I meet your intentions)

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  • $\begingroup$ Thanks for the elaborate answer. Makes sense. Indeed it is derived from Collatz :). $\endgroup$ – Catalin Neacsu Dec 17 '19 at 9:58

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