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Writing $3\cdot 2^a=(b-1)(b+1)$ , we see that either one of the numbers $b-1$ and $b+1$ must be a power of $2$ ; the other must be a power of $2$ multiplied by $3$ . If $b$ is even, we must have $(b-1)(b+1)=1\cdot 3$ , giving $(a,b)=\boxed{(0,2)}$ . If $b$ is odd, either $b+1=6$ , giving $(a,b)=\boxed{(3,5)}$ , or $b-1=6$ , giving $(a,b)=\boxed{(4,7)}$ .

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The possible pairs are (0,2),(3,5),(4,7)

We write $3\cdot 2^a=(b-1)(b+1)$ , so we must have that one of $b-1$ and $b+1$ is a power of $2$ , and the other is $3$ times a power of $2$ . This means that $(b-1,b+1)=(3\cdot 2^x, 3\cdot 2^x\pm 2)$ . We'll do casework to find all solutions.

Case 1: $3\cdot 2^x+2$

If this were the case, then we note that we must have $3\cdot 2^x+2=2^{x+2}$ since $2<3<2^2$ , and the other term must be a power of $2$ . Solving gives $3\cdot 2^x+2=2^{x+1}+2^x+2=2^{x+2}$ $2^x+2=2^{x+2}-2^{x+1}=2^{x+1}$ $2=2^{x+1}-2^x=2^x \implies x=1$ which gives the solution $(b-1,b+1)=(3\cdot 2,3\cdot 2+2)=(6,8) \implies \boxed{(a,b)=(4,7)}$ .

Case 2: $3\cdot 2^x-2$

This time we must have that $3\cdot 2^x-2=2^x$ or $3\cdot 2^x-2=2^{x+1}$ due to bounding on $3$ again. If the former were the case, we get $2^{x+1}-2=0 \implies x=0$ which gives the solution $(b-1,b+1)=(1,3) \implies \boxed{(a,b)=(0,2)}$ . If the latter were the case we get $3\cdot 2^x-2=2^{x+1}+2^x-2=2^{x+1}$ $2^x-2=0\implies x=1$ which gives the solution $(b-1,b+1)=(4,6)\implies \boxed{(a,b)=(3,5)}$ . Hence, the solution is $0+2+3+5+4+7=\boxed{21}$ .

Rewriting it as $3×2^a=(b+1)(b-1)$ shows we need to look for two numbers, two apart, picking one from the set {3,6,12,24,48,..} and the other from the set {1,2,4,8,16,32,...}. We have the following solutions:

• $1,3 \Rightarrow b=2, a=0$
• $4,6 \Rightarrow b=5, a=3$
• $6,8 \Rightarrow b=7, a=4$

Summing up all a's and b's : $2+0+5+3+7+4=\boxed{21}$