Let:
- $n\ge 4, x, a, b$ be integers
- gcd$(c,d)$ be the greatest common divisor of $c$ and $d$
- gcd$(x,6)=1$
- $p_1, p_2, \dots, p_n$ be positive integers
- $q_1, q_2, \dots, q_n$ be positive integers
- $P_i = \sum\limits_{k=1}^i p_k$
- $Q_i = \sum\limits_{k=1}^i q_k$
Question:
It seems to me that for a large enough $n$, $x[3^{n-1}a + \sum\limits_{i=1}^{n-1}(3^{n-1-i}2^{P_i})]$ cannot have the form: $3^{n-1}b + \sum\limits_{i=1}^{n-1}(3^{n-1-i}2^{Q_i})$.
I am not clear how to prove this. Here's my thinking:
- I suspect that the product of $x$ and $3^{n-1}a + \sum\limits_{i=1}^{n-1}(3^{n-1-i}2^{P_i})$ cannot have this form for $n \ge 4$.
- I suspect the product can have this form when $n=2$.
- I have found that the product can have this form when $x$ is a power of $2$ such as $1, 2, 4, \dots$
Could someone help me to find an argument to show that the product cannot have this form, a counter example to show that it can, or an explanation for why the problem is difficult to resolve one way or the other.
Edit:
Note: Thanks to John Omielan for catching a mistake in my edit. I have updated this edit.
I believe that I have found a counter example when there exists integers $m,b$ where $x = 2^m + 3^{n-1}b$.
Let:
- $T = 3^{n-1}a + \sum\limits_{i=1}^{n-1}(3^{n-1-i}2^{P_i})$
Then: $$xT = 3^{n-1}(a2^m+bT) + \sum\limits_{i=1}^{n-1}(3^{n-1-i}2^{P_i+m})$$
Would this be the only counter example? If no such $b$ exists for any value $m$, does that imply that the product would not have the form?
Edit 2:
Note: I have updated the detail of the problem to clarify that in the form $a$ means any integer. From John's comment, it becomes clear that this opens up the number of counter examples beyond the one in edit 1.