Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For any number $n \gt 1$ and all of its prime divisors $d_1, d_2, ...$ s.t.

$d_i \equiv 1 \pmod 3$ for each $i$

Show that the euler phi function $\phi(x) = 2n$ has no natural number solution.

share|improve this question
    
You mean: numbers $n$ such that all prime divisors of $n$ are congruant to $1$ modulo $3$? –  Marc van Leeuwen Oct 17 '12 at 10:16

1 Answer 1

Let $x = \prod_{i=1}^n p_i^{\alpha_i}$ ($p_i$ the distinct prime factors of $x$) be given, we have \[ \phi(x) = \prod_{i=1}^n p_i^{\alpha_i - 1}(p_i - 1). \] So in order to have $\phi(x) = 2n$ with $n$ odd (as above) and $p_i - 1$ being even for all primes but 2, we see that $x$ must be of the form $x = 2^\alpha p^\beta$ with $\alpha \in \{0,1\}$. Then \[ \phi(x) = p^{\beta - 1}(p-1). \] This equals $2n = 2d_1\ldots d_r$ if (after reordering the $d_i$) $p-1 = 2d_1$ (or $=2$), $p^{\beta - 1} = d_2 \cdots d_r$. So $p = d_2 = \cdots = d_r$, and $2d_1 + 1 = p$. But $d_1 \equiv 1\pmod 3$, so $2d_1 + 1 \equiv 0 \pmod 3$. But $p$ is prime, so we must have $p-1 = 2$, all $d_i$ equval $p$. But then $p = 3$, which contradicts $p = d_1 \equiv 1 \pmod 3$. So $n$ can't have any prime divisors, contradicting $n > 1$.

So there is no such $x$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.