The integers $n$ with $\phi(n)=2$ are $3,4,6$

1. Let \(n=\prod_{i=1}^r p_i^{\alpha_i}\) be the prime factorization. Then \[ \phi(n)=\prod_{i=1}^r \phi(p_i^{\alpha_i})=2,\qquad \phi(p_i^{\alpha_i})=p_i^{\alpha_i-1}(p_i-1)\in\mathbb Z_{>0}. \]
2. Every factor \(p_i^{\alpha_i-1}(p_i-1)\) equals \(1\) or \(2\), and exactly one factor equals \(2\).
3. If one factor equals \(2\) for an odd prime \(p_i\), then \(\alpha_i=1\) and \(p_i-1=2\), so \(p_i=3\).
4. If one factor equals \(2\) for \(p_i=2\), then \(2^{\alpha_i-1}(2-1)=2\), hence \(\alpha_i=2\) so \(2^2\mid n\).
5. Any remaining factors equal \(1\), so \(\phi(p_i^{\alpha_i})=1\). This forces \(p_i=2\) and \(\alpha_i=1\).
6. Therefore
   • \(n=3\) if the factor \(2\) comes from \(3\) and no extra \(2\)-factor appears;
   • \(n=4\) if it comes from \(2^2\);
   • \(n=6\) if both \(3\) and one factor of \(2\) appear. No other primes can appear because they would contribute a factor \(>2\).
7. Conversely, \(\phi(3)=\phi(4)=\phi(6)=2\).

So the only integers with \(\phi(n)=2\) are exactly \(3,4,6\).