# Can the uniqueness of the solution be proved?

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

• Is $3^0$ intended ? – Yves Daoust Dec 13 '19 at 10:00
• Hi Yves, it is a typo. It should be a 1 to match the pattern. I fixed that. Thanks for the comment. – Catalin Neacsu Dec 14 '19 at 16:13
• 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. – Catalin Neacsu Dec 14 '19 at 16:17
• 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) – Gottfried Helms Dec 15 '19 at 23:38

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