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Prove that $\varphi(m)+ \tau(m)\leqslant m+1$ where $m\in \mathbb N$

I wrote $m:=p_1^{\alpha_ 1}....p_s^{\alpha _s}$

$$\varphi(m)=p_1^{\alpha_1}(p_1-1)...p_s^{\alpha_s}(p_s-1)$$

$$\tau(m)=(\alpha_1+1)...(\alpha_s+1)$$

Then $\varphi(m)+ \tau(m)=\prod\limits_{i=1}^{s}p_i^{\alpha_i-1}(p_i-1)+\sum\limits_{i=1}^{s}(\alpha_i+1)\overset{?}{\leqslant}\prod\limits_{i=1}^{s} p_i^{\alpha_i}+1$

and I'm stuck here, how can I prove that $\prod\limits_{i=1}^{s}p_i^{\alpha_i-1}(p_1-1)+\sum\limits_{i=1}^{s}(\alpha_i+1)\overset{?}{\leqslant}\prod\limits_{i=1}^{s} p_i^{\alpha_i}+1$

or maybe is there an easy way to prove that $\varphi(m)+ \tau(m)\leqslant m+1$

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The divisors of m are not relatively prime to m, except for 1. Therefore each number from 1 to m is counted at most once, except 1, which is counted twice, for a maximum of m+1.

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  • 1
    $\begingroup$ I had to read your first sentence a few times before I realized that the missing word must be “prime” … $\endgroup$ – Harald Hanche-Olsen Apr 12 '16 at 12:56
  • $\begingroup$ Thank. Fixed. That's what I get for entering it on my phone. $\endgroup$ – marty cohen Apr 12 '16 at 13:59
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$$ m = \sum_{d \mid m} \phi(d) = \phi(m) + \sum_{d \mid m, d < m} \phi(d) \ge \phi(m) + \sum_{d \mid m, d < m}1 \ge \phi(m) + \tau(m) -1 $$

As an aside, we have equality in $\varphi(m)+ \tau(m)\leqslant m+1$ iff every proper divisor $d$ has $\phi(d)=1$, which happens iff the divisors of $n$ are $1,n$ or $1,2,n$, and this happens iff $n$ is prime or $4$.

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  • $\begingroup$ Nice use of the equation involving $\phi$. $\endgroup$ – marty cohen Apr 13 '16 at 6:13
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Fix $m\in\mathbb{N}$ and let $A=\{a\in\mathbb{N}:a\leq m,\gcd{(a,m)}=1\},$ $B=\{a\in\mathbb{N}:a\mid m\}$. Now $\lvert A \rvert = \varphi(m)$ and $\lvert B \rvert = \tau(m)$. Hence $$\varphi(m)+\tau(m) = \lvert A \rvert + \lvert B \rvert = \lvert A \cup B \rvert + \lvert A\cap B \rvert.$$

But now note that $A\cup B\subseteq\{1,2,\ldots,m\}$ and $A\cap B = \{1\}$, so $\lvert A\cup B \rvert \leq m$ and $\lvert A\cap B \rvert = 1$. The result follows.

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  • $\begingroup$ Essentially the same as mine, but definitely more mathy. $\endgroup$ – marty cohen Apr 13 '16 at 6:14

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