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(A) Given the equation: $(a-8)^2-(b-7)^2=5$This can be written as: $(a-8-b+7)(a-8+b-7)=5$$\Rightarrow (a-b-1)(a+b-15)=5$Since $a$ and $b$ are integers

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$20$

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(A) Given the equation: $(a-8)^2-(b-7)^2=5$This can be written as: $(a-8-b+7)(a-8+b-7)=5$$\Rightarrow (a-b-1)(a+b-15)=5$Since $a$ and $b$ are integers,$(a-b-1)$ and $(a+b-15)$ must be integer factors of $5$. The possible pairs of factors $(x, y)$ such that $xy=5$ are $(1, 5), (5, 1), (-1, -5), (-5, -1)$.Case $1$: $a-b-1=1$ and $a+b-15=5 \Rightarrow a-b=2$ and $a+b=20$. Adding gives $2a=22 \Rightarrow a=11$,then $b=9$.Case $2$: $a-b-1=5$ and $a+b-15=1 \Rightarrow a-b=6$ and $a+b=16$. Adding gives $2a=22 \Rightarrow a=11$,then $b=5$.Case $3$: $a-b-1=-1$ and $a+b-15=-5 \Rightarrow a-b=0$ and $a+b=10$. Adding gives $2a=10 \Rightarrow a=5$,then $b=5$.Case $4$: $a-b-1=-5$ and $a+b-15=-1 \Rightarrow a-b=-4$ and $a+b=14$. Adding gives $2a=10 \Rightarrow a=5$,then $b=9$.The vertices are $(11, 9), (11, 5), (5, 5), (5, 9)$.The length of the sides are the distances between adjacent vertices:Length $L = \sqrt{(11-11)^2+(9-5)^2} = 4$Width $W = \sqrt{(11-5)^2+(5-5)^2} = 6$Perimeter $= 2(L+W) = 2(4+6) = 20$.

All the vertices of a rectangle are of the form $(a, b)$ with $a, b$ integers satisfying the equation $(a-8)^2-(b-7)^2=5$. Then,the perimeter of the rectangle is

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