A mind eating algebra problem

Given that $\begin{cases} ab+a+b = 524 \\ bc+b+c = 146 \\ cd+c+d = 104 \end{cases}$ . we note that all $a$ , $b$ , $c$ and $d$ are even; also that $\begin{cases} (a+1)(b+1) = 525 &...(1) \\ (b+1)(c+1) = 147 &...(2) \\ (c+1)(d+1) = 105 & ...(3) \end{cases}$

$\implies \dfrac {(1)}{(2)}: \dfrac {a+1}{c+1} = \dfrac {25}7$ , Since $c$ is even the smallest possible value is $c=6$ for $a=24$ to be an integer. The next value $c=20$ , then $a=524$ which is too big.

Now consider

$\begin{aligned} \frac {(1)\times(3)}{(2)}: \quad ({\color{#3D99F6}a}+1)(d+1) & = \frac {525 \times 195}{147} = 375 & \small \color{#3D99F6} \text{Note that }a=24 \\ 25(d+1) & = 375 \\ d+1 & = 15 \\ \implies d & = 14 \end{aligned}$

$\implies a-d = \boxed{10}$ .

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Checking: From $(2):$ we get $b=20$ , then $abcd = 24 \times 20 \times 14 \times 6 = 2^7\times 3^2 \times 5 \times 7 = 8!$ .