Daylight problem

$3|({ b }^{ 2 }-1)$ and $2|({ b }^{ 2 }-1)$ , so $b$ is odd and $b\equiv 1(mod\quad 3)$ or $b\equiv 2(mod\quad 3)$ .

In the latter case, $b=6k-1$ , thus,

$3\cdot { 2 }^{ a }+1=36{ k }^{ 2 }-12k+1\\ \\ { 2 }^{ a }=12{ k }^{ 2 }-4k$

If $a=0$ , then $k=1/2$ or $k=-1/3$ , so $b=2$ or $b=-2$ , but we discart the second option, because $b>0$ .

If $a=1$ or $a=2$ , then $k$ is not rational. If $a>2$ , then

${ 2 }^{ a-2 }=k(3k-1)$ , but $(3k-1)$ is even only when $k$ is odd, and we want $k={ 2 }^{ m }$ for some $m<a-2$ , which is even, therefore, $m=0$ and $k=1$ , hence, $b=5$ and $a=3$ .

In the former case, $b=6k+1$ , thus,

$3\cdot { 2 }^{ a }+1=36{ k }^{ 2 }+12k+1\\ { 2 }^{ a }=12{ k }^{ 2 }+4k$

If $a\le 2$ , we have the same cases that before. If $a>2$ , then

${ 2 }^{ a-2 }=k(3k+1)$ , but $(3k+1)$ is even only when $k$ is odd, and we want $k={ 2 }^{ m }$ for some $m<a-2$ , which is even, therefore, $m=0$ and $k=1$ , hence, $b=7$ and $a=4$ .

The sum of all possible values for $a$ and $b$ that satisfy the given conditions is

$2+0+5+3+7+4=21$