Divisor as the Numerator of a Quotient

Given that $\dfrac {a-1}b = \dfrac b{c^4}$ , $\implies b^2 = c^4(a-1) \implies b = c^2\sqrt{a-1}$ . This implies that $a-1$ must be a perfect square. Let $n^2 = a-1$ . Since $0 < a < 100$ , $1 \le n \le 9$ .

Then we have $a=n^2 +1$ and $b = n c^2$ . Let $s = b-a-c = nc^2 -n^2 - 1 - c$ . We note that for a particular $n$ , $s$ increases with $c$ . Implying that $s_{\max} = nc_{\max}^2 - c_{\max} - n^2 - 1$ , where $c_{\max} = \left \lfloor \sqrt{\dfrac {b_{\max}}n} \right \rfloor = \left \lfloor \sqrt{\dfrac {99}n} \right \rfloor$ . Finally,

$\begin{aligned} s_{\max}(1) & = \left \lfloor \sqrt{\frac {99}1} \right \rfloor^2 - \left \lfloor \sqrt{\frac {99}1} \right \rfloor - 1 -1 = 81 - 9 - 2 = 70 \\ s_{\max}(2) & = 2\left \lfloor \sqrt{\frac {99}2} \right \rfloor^2 - \left \lfloor \sqrt{\frac {99}2} \right \rfloor - 4 -1 = 2(7^2) - 7 - 5 = 86 \\ s_{\max}(3) & = 3\left \lfloor \sqrt{\frac {99}3} \right \rfloor^2 - \left \lfloor \sqrt{\frac {99}3} \right \rfloor - 9 -1 = 3(5^2) - 5 - 10 = 60 \end{aligned}$

Similarly, we have $s_{\max} (4) = 43$ , $s_{\max} (5) = 50$ , $s_{\max} (6) = 55$ , $s_{\max} (7) = 10$ , $s_{\max} (8) = 4$ , and $s_{\max} (9) = -4$ . Therefore the highest value of $b-a-c = 98-5-7 = \boxed{86}$

take $a=5,b=18,c=3$ to have $b-a-c=18-5-3=10 > 4$ to show that there is something wrong?