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Let

  • $\varphi(n)$ be the totient-function
  • $\sigma(n)$ be the divisor-sum-function

It is clear that every prime number $n$ is the arithmetic mean of $\varphi(n)$ and $\sigma(n)$, in other words , the equality $$\varphi(n)+\sigma(n)=2n$$ holds.

Can the given equality hold for a composite number $n$ ?

Upto $10^8$, I did not find an example. I conjecture that the equality can only hold for primes or $n=1$. How can I prove it ?

Peter
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3 Answers3

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In other terms, we want that

$$ \prod_{p\mid n}\left(1-\frac{1}{p}\right)+\prod_{p\mid n}\left(1+\frac{1}{p}+\ldots+\frac{1}{p^{\nu_p(n)}}\right) = 2 $$ but the LHS cannot be an integer if $\nu_p(n)>1$, so it is enough to restrict our attention to square-free numbers and to the solutions of $$ \prod_{p\mid n}\left(1-\frac{1}{p}\right)+\prod_{p\mid n}\left(1+\frac{1}{p}\right) = 2. $$ There are some solutions only if $\omega(n)=0$ or $\omega(n)=1$.
If $\omega(n)\geq 2$, the LHS is at least as large as $$ 2+\sum_{\substack{p,q\mid n \\ p\neq q}}\frac{1}{pq} > 2.$$

Jack D'Aurizio
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    Alternatively note that $\underbrace{\prod(1-\frac1p)}{=:P_1}+\underbrace{\prod(1+\frac1p)}{=:P_2}=2$ if the products have at most one factor, whereas each further factor turns $P_1+P_2$ into $P_1(1-\frac1p)+P_2(1+\frac1p)=P_1+P_2+\frac1p(P_2-P_1)>P_1+P_2$ because $P_2>P_1$; hence $P_1+P_2\ge2$ with equality iff there is at most one $p$. And adding $\ldots+\frac1{p^{\nu_p(n)}}>0$ into any factor pf $P_2$ makes the result $>2$ as well. – Hagen von Eitzen Aug 08 '18 at 02:20
  • @HagenvonEitzen Why can we conclude in Jack's answer that the first expression on the left side cannot be an integer if we do not have a squarefree number ? Does he perhaps just mean that it must exceed $2$ ? – Peter Aug 08 '18 at 22:12
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The answer is indeed "No". Let $n=p_1^{a_1}\cdots p_k^{a_k}$. Then $$\sigma(n)=(1+p_1+\cdots+p_1^{a_1})\cdot(1+p_2+\cdots+p_2^{a_2})\cdots(1+p_k+\cdots+p_k^{a_k})\\\varphi(n)=(p^{a_1}_1-p^{a_1-1}_1)\cdots (p_k^{a_k}-p_k^{a_k-1})$$

Now, if you expand the product in $\varphi(n)$ and take a look at the "monomials" $p_1^{h_1}\cdots p_k^{h_k}$ that appear in it, you'll notice that:

  1. they all have coefficient either $+1$ or $-1$ (the case $0$ is not considered because we're dealing with the ones that appear in the expansion)

  2. they are a subset of the monomials that appear in the expansion of $\sigma(n)$

  3. $p_1^{a_1}\cdots p_k^{a_k}$ appears with coefficient $+1$

  4. if $k\ge 2$, there is at least one monomial $M_n$ that appears with coefficient $+1$ other than the one mentioned in (3) - namely, $M_n=p_1^{a_1-1}p_2^{a_2-1}\prod_{j=3}^k p_j^{a_j}$.

Since all the monomials in $\sigma(n)$ have coefficient $+1$, the ones that have $-1$ in $\varphi(n)$ cancel out upon sum with $\sigma(n)$. Therefore we obtain that $$\sigma(n)+\varphi(n)\ge 2\cdot\sum\{\text{monomials that appear with coefficient }+1\text{ in }\varphi(n)\}$$

As we've observed in (3) and (4), if $k\ge 2$ we have that $RHS\ge 2n+2M_n> 2n$. On the other hand, if $n=p^a$, then $$\sigma(p^a)+\varphi(p^a)=2p^a+0\cdot p^{a-1}+p^{a-2}+\cdots +1$$ which is equal to $2p^a$ if and only if $a<2$.

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Noticed that it's equivalent to prove that $$ \sigma\left(n\right)-n\ge n-\varphi \left(n\right) $$ Noticed that $RHS$ means how many $i$ and $n$ is not relatively prime where $i\in\left[1,n\right]$ $$ RHS=\sum_{i=1}^n\left[\gcd\left(i,n\right)\neq1\right] $$ So $$ RHS\le \sum_{k=2,k \mid n}^n\sum_{i=1}^n\left[k \mid i\right]=\sum_{k=2,k \mid n}^n\dfrac n k=LHS $$

songhongyi
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