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Let $n_l = 7 \cdot 10^{l+1}-2 = 699\ldots 998$. I want to know if there is an $l$ such that $n_l$ can occur as the value of the Euler totient function $\varphi(n)$. For given $l$, it is easy to check if this can happen (and I showed that $n_l \notin \varphi(\mathbb{N})$ for $l \leq 100$). For general $l$, I have no idea how I could show this (if it is true). What I tried so far:

  • Suppose that $\varphi(m) = n_l$. Since $n_l \equiv 2 \bmod 4$ we must have $m = p^k$ or $m=2p^k$ for an odd prime $p$, hence $n_l = p^{k-1} (p-1)$.
  • The polynomial $x^2-x-n_l$ has no rational roots, since its discriminant is irrational, hence we must have $k \geq 2$.
  • Building on the idea of the point above, consider the polynomial $x^{k+1}-x^k-n_l$. Is there anything known about roots of this polynomial or its Galois group? I am aware of the rational root theorem, but this does only help for specific $l$.

If considering the polynomial does not help: Does anyone have another idea how one could handle this question?

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  • $\begingroup$ Since $n_l \equiv 2 \mod 3$, the prime $p \equiv 2 \mod 3$, and $k$ is even. $\endgroup$ Jul 10, 2017 at 15:56

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