Euler's Totient Function Equations (Problem $1$ )

Since Euler's totient function value is $8$ it is likely that one of the operants $x-1, x, ...$ is a multiple of $3$ and $5$ . Checking $15$ , we have $\phi(15) = 15 \times \dfrac 23 \times 45 = 8$ . The operant can also be a power of $2$ and $2^4 = 16$ satisfies the equation $\phi (16) = 16 \times \dfrac 12 = 8$ . Therefore $x=\boxed{16}$ , so the $x-1 = 15$ . And

$\begin{aligned} \phi(x+8) & = \phi (24) = 24 \times \frac 12 \times \frac 23 = 8 \\ \phi(2x-2) & = \phi (30) = 30 \times \frac 12 \times \frac 23 \times 45 = 8 \end{aligned}$