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Determine all n for which [itex]\phi(n) = n -2[/itex].

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Now it seems that the only time this will work is for n = 4. However, I haven't any idea of how to prove (or justify) this. I have thought about working primes and composites, since we know [itex]\phi(p) = p-1[/itex] for all primes p.

Some things we know:

If (m,n)=1, then [itex]\phi(mn) = \phi(m)\phi(n)[/itex].

If [itex]m=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}[/itex] is the prime factorization of m, then:

[tex]\phi(m) = \prod_{i=1}^k \left(1- \frac{1}{p_i}\right)[/tex]

Any hints/ideas? Thanks!

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# Homework Help: Number Theory: Euler's Phi Function

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