33 In triangle $A B C, A B=\left(b^{2}-1\right) \mathrm{cm}, B C=a^{2} \mathrm{~cm}$ and $A C=2 a \mathrm{~cm}$, where $a$ and $b$ are positive integers greater than 1 . Find the value of $a-b$.
33. Answer: 0
Using Triangle Inequality, $a^{2}+2 a>b^{2}-1 \Rightarrow(a+1-b)(a+1+b)>0$. Since $a+$ $1+b>0 \Rightarrow a+1-b>0 \Rightarrow a-b \geq 0$. Using Triangle Inequality again, $2 a+b^{2}-$ $1>a^{2} \Rightarrow(a-1-b)(a-1+b)0 \Rightarrow a-1-b<0 \Rightarrow a-b$ $\leq 0$. Hence $a-b=0$.