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I am interested in finding the solutions to the equation below: $$\lambda = \frac{2^c+3b}{2^d-3^a}$$ $$\lambda, a, b, c, d \in \Bbb Z^+$$ $$d>a,c$$

Edit: My original version was actually: $$\frac{3^a(\lambda) + 3(3b+2^c)+2^d}{2^e}= \lambda$$ $$e>a,c,d$$

Furthermore, when looking at this original equation, it became apparent that this was actually:

$$\frac{3^a \lambda+3^a b +\sum_{i=o}^{a-1} 3^i 2^{j_i}}{2^c}=\lambda$$ Where $j_i \in \Bbb Z^+ < d-1$. At this point, I am lost as to how I should attempt to solve this.

Are there solutions to the original equations, other than $1$ and $2$?

Can anyone find large solutions? ($a,b,c,d > 100$)? I am only really interested in the large solutions.

Also, is there a technique to find solutions to this as I am really not sure how to go about this.

Roskiller
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1 Answers1

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I would recommend to describe the original problem before the generalisation, because in its current form without any restrictions the solution is straightforward.

The equation in positive integers $a,b,c,d,\lambda$ $$ \lambda = \frac{2^c + 3 b}{2^d - 3^a} $$ can immediately be rewritten as $$ 3 b = \lambda 2^d - 2^c - \lambda 3^a $$ So the only constraint on the integers (using $d>c$) we find is that $3 | \lambda 2^d - 2^c = 2^c (\lambda 2^{d-c} - 1)$. This results in either a solution with $\lambda \equiv 1~\text{mod}~3$ and $d-c \equiv 0~\text{mod}~2$ or $\lambda \equiv 2~\text{mod}~3$ and $d-c \equiv 1~\text{mod}~2$. There are no solutions with $\lambda$ being a multiple of 3.

For any choice of positive $a$,$c$, and $\lambda$ (not a multiple of 3) either all even or odd values of $d > \max\left(c,\frac{\log 3^a + 2^c/\lambda}{\log 2}\right)$ will be enough to guarantee an integer solution for $b$. Note that the constraint $d>a$ is automatically satisfied.

The solution for the original problem is more or less the same. Rewriting it gives that $$ 9 b = \lambda 2^e - 2^d - \lambda 3^a - 3 \times 2^c $$ Assuming $a,c,d,e$ are positive the RHS needs to be a multiple of 3, hence $\lambda \equiv 1(\text{or } 2)\text{ mod }3$ and $e - d \equiv 0 (\text{or }1) \text{ mod } 2$.

It gets a little bit more complicated, because the RHS needs to be a multiple of 9. This means that for each of the 4 terms you need to determine what its remainder modulo 9 will be and how that is related to the values of $c,d,e$ modulo 6 whereas for $\lambda$ it is modulo 9. Also you need to distinguish the cases $a=1$ and $a \geq 2$. After this analysis, which is not too difficult but just many cases, you can choose any suitable combination of $a,c,d,e,\lambda$ with the only additional restriction that $e$ should be large enough to make $b$ positive.

Ronald Blaak
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  • Can you give an example $>10$ as I don't quite follow what you mean? – Roskiller Aug 16 '18 at 17:29
  • Sure: $a=17$, $c=23$, $\lambda=53$, then $d$ must be even and larger than $\max(23,26.9461..)$. For $d=28$ you find $b=2458087307$, for $d=30$ you get $b=16685166475$, etc. – Ronald Blaak Aug 16 '18 at 17:39
  • Thanks for showing the solution to those two, but how could I go about the last one, as that is the one I am most interested in. – Roskiller Aug 19 '18 at 14:03
  • could you perhaps say something about the last equation I have added as it seem more complex than the others. – Roskiller Aug 27 '18 at 16:04
  • No it isn't more complex. A solution in the same style can be formulated, provided you understand the "solution" to the other problems. What is an issue, is that it stops being a clean problem, because you ask for solutions to an equation with many unknown variables that does not appear to lead anywhere. It looks very much like "let me ask a very complicated question". Why is this interesting? Where does this appear? What is the actual problem you like to solve? What are the restrictions? How much of it can you solve yourself and what is the problem you get stuck with? – Ronald Blaak Aug 27 '18 at 17:10