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Let $\sigma_1$ be the classical sum-of-divisors function.

Let $$D(x) = 2x - \sigma_1(x)$$ be the deficiency of $x$.

Let $\varphi(x)$ denote the Euler totient.

Here is my question:

For what numbers $x$ is $D(x) = 2x - \sigma_1(x)$ equal to $\varphi(x)$?

Trivially, $x=1$ answers my question. But are there any others?

Jose Arnaldo Bebita Dris
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    Only $x=1$ and $x=p$ for some prime natural number $p$ are the equality case. Your other thread on a "sharp" bound of $D(x)$ should provide a clue on why that is. You can first show that $x$ must be squarefree, and then $x$ can have at most one prime factor. – Batominovski Aug 28 '16 at 15:49
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    $\sigma(n) = \sum_{d | n} d$ and $\phi(n) = \sum_{d | n} d \mu(n/d)$ so $\frac{\sigma(n)+\phi(n)}{2} = \sum_{d | n} d \frac{1+\mu(n/d)}{2} \ge n$ and it is $= n$ iff $\frac{1+\mu(n/d)}{2} = 0$ whenever $d > 1$ i.e. iff $n$ is prime – reuns Aug 28 '16 at 17:25

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