It is it possible to always find arbitrarily large solutions to this equation?

Question

I have been playing around with Collatz sequences a lot recently, and so now I want something slightly different but not too different, if you get what I mean.

Solving puzzles online doesn't quite fit the bill and so I came up with this:

Is it possible to always find infinite arbitrarily large positive integer solutions $a,b,c \in \Bbb Z^+$, such that:

$$2^a \,\big | \,3b + 2^c$$

Edit: This was the generalised version of my original question which was actually: $$2^a|3^b+3c+2^d$$

An answer to (Edit: Both these) would be greatly appreciated as my approaches, (modular arithmetic, algebra, graphing) have been quite fruitless. Thanks.

Answer 1

Yes: for instance $(a,b,c)=(k,2^{k-2},k-2)$ for $k\ge 3$ are some solutions. Some others are:

• $(a,5\cdot 2^{c},c)$ for any $5\le a\le c+4$;

• $(a,21\cdot 2^c,c)$ for any $7\le a\le c+6$;

• $\vdots$

Answer 2

We know that $2^k \equiv 1 \pmod 3$ when $k$ is even and $2$ when $k$ is odd. You can choose any $a,c$ of the same parity with $a \ge c+2$ and $2^a-2^c$ will be divisible by $3$, so set $b=\frac 13(2^a-2^c)$. This is an expansion of Saucy O'Path's answer.

Another family of solutions is $c \ge a, b=2^a$. There are many more.

Answer 3

By examination, whenever $b=n \cdot 2^d$ and $c,d \ge a$, $3b+2^c$ is divisible by $2^a$. So to the question posed, are there infinite solutions, the answer is yes.