Euler Phi-Function

To start with, note that, since $\phi(n)$ is either 1 or an even number, n must be even.

Remember that: $ \phi \left( n \right) = n \cdot \prod\limits_{\left. p \right|n} {\left( {1 - \tfrac{1} {p}} \right)} $ (the products go over the prime numbers p dividing n)

Thus: $ \phi \left( n \right) = \tfrac{n} {3} \Leftrightarrow \prod\limits_{\left. p \right|n} {\left( {1 - \tfrac{1} {p}} \right)} = \tfrac{1} {3} \Leftrightarrow 3 \cdot \prod\limits_{\left. p \right|n} {\left( {p - 1} \right)} = \prod\limits_{\left. p \right|n} p $

Now since 3 is a prime number, and $ 3\left| {\prod\limits_{\left. p \right|n} p } \right. $ it follows that 3 is one of the prime factors in the product. Thus: $2\cdot \prod\limits_{\left. p \right|n;p \ne 3} {\left( {p - 1} \right)} = \prod\limits_{\left. p \right|n;p \ne 3} p $ and 2|n thus: $ \prod\limits_{\left. p \right|n;p \ne 3;p \ne 2} {\left( {p - 1} \right)} = \prod\limits_{\left. p \right|n;p \ne 3;p \ne 2} p $ and this is only possible if there are no other prime factors.

Thus: $ n = 2^\alpha \cdot 3^\beta $ with $\alpha,\beta\in \mathbb{Z}^+$

And conversely this number satisfies the condition for any $\alpha,\beta\in \mathbb{Z}^+$