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Math Help - Euler Phi function property

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    Euler Phi function property

    Hello everyone!

    I'm working on this problem:

    Q: Find all n such that \varphi\!\left(n\right)=n-2.

    Here's my work:

    Consider the following cases:

    Case 1: n is prime.

    In n is prime, then \varphi\!\left(n\right)=n-1. However, if this is supposed to satisfy the above requirement, then we must have n-1=n-2 which is not possible for any n.

    Case 2: n=pq, with p and q prime.

    We have \varphi\!\left(pq\right)=\varphi\!\left(p\right)\v  arphi\!\left(q\right)= (p-1)(q-1). To satisfy the above equality, we must have (p-1)(q-1)=pq-1\implies pq-p-q+1=pq-2\implies p+q=3 which is not possible since 1 is not a prime.

    Case 3: n=p^2, with p prime.

    We have \varphi\!\left(p^2\right)=p^2-p. To satisfy the above equality, we must have p^2-p=p^2-2\implies p=2. Therefore n=4 is a solution.

    Now, I tried other cases that failed (like p^2q, p^2q^2, p^3q). I believe that n=4 is the only solution, but my main question is when do I know where to stop with all these different cases?

    Any help is appreciated!
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  2. #2
    Super Member PaulRS's Avatar
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    Note that if n>2 then \phi(n) is even - that is simple, if (k,n)=1 then we also have (n-k,n)=1 ... -.

    That implies that n is even -by your equation-, so if n>4, n has at least 3 divisors not coprime to it (n, 2 and n/2) (*), but then \phi(n)<n-2 a contradiction!.
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