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:

$\map \phi n = \map \phi {n + 2}$

Proof

From Euler Phi Function of Prime:

$\map \phi 7 = 7 - 1 = 6$

From Euler Phi Function of Prime Power:

$\map \phi 9 = \map \phi {3^2} = 2 \times 3^{2 - 1} = 6 = \map \phi 7$

From the corollary to Euler Phi Function of Prime Power:

$\map \phi 8 = \map \phi {2^3} = 2^{3 - 1} = 4$

From Euler Phi Function of Integer:

$\map \phi {10} = \map \phi {2 \times 5} = 10 \paren {1 - \dfrac 1 2} \paren {1 - \dfrac 1 5} = 10 \times \dfrac 1 2 \times \dfrac 4 5 = 4 = \map \phi 8$

$\blacksquare$

Sources

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $7$