Successive Solutions of Phi of n equals Phi of n + 2

Theorem
Let $\phi$ denote the Euler $\phi$ function. $7$ and $8$ are two successive integers which are solutions to the equation:
 * $\phi \left({n}\right) = \phi \left({n + 2}\right)$

Proof
From Euler Phi Function of Prime:
 * $\phi \left({7}\right) = 7 - 1 = 6$

From Euler Phi Function of Prime Power:
 * $\phi \left({9}\right) = \phi \left({3^2}\right) = 2 \times 3^{2 - 1} = 6 = \phi \left({7}\right)$

From the corollary to Euler Phi Function of Prime Power:
 * $\phi \left({8}\right) = \phi \left({2^3}\right) = 2^{3 - 1} = 4$

From Euler Phi Function of Integer:
 * $\phi \left({10}\right) = \phi \left({2 \times 5}\right) = 10 \left({1 - \dfrac 1 2}\right) \left({1 - \dfrac 1 5}\right) = 10 \times \dfrac 1 2 \times \dfrac 4 5 = 4 = \phi \left({8}\right)$