Theorem: Integers whose Phi times Divisor Count equal Divisor Sum

The positive integers whose Euler $\phi$ function multiplied by its  divisor count function equals its divisor sum are:
:$1, 3, 14, 42$
{{OEIS|A104905}}

Proof:
{{begin-eqn}}
{{eqn | l = \map \phi 1 \map {\sigma_0} 1
      | r = 1 \times 1
      | c = {{EulerPhiLink|1}}, {{DCFLink|1}}
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | r = \map {\sigma_1} 1
      | c = {{DSFLink|1}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map \phi 3 \map {\sigma_0} 3
      | r = 2 \times 2
      | c = {{EulerPhiLink|3}}, {{DCFLink|3}}
}}
{{eqn | r = 4
      | c = 
}}
{{eqn | r = \map {\sigma_1} 3
      | c = {{DSFLink|3}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map \phi {14} \map {\sigma_0} {14}
      | r = 6 \times 4
      | c = {{EulerPhiLink|14}}, {{DCFLink|14}}
}}
{{eqn | r = 24
      | c = 
}}
{{eqn | r = \map {\sigma_1} {14}
      | c = {{DSFLink|14}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map \phi {42} \map {\sigma_0} {42}
      | r = 12 \times 8
      | c = {{EulerPhiLink|42}}, {{DCFLink|42}}
}}
{{eqn | r = 96
      | c = 
}}
{{eqn | r = \map {\sigma_1} {42}
      | c = {{DSFLink|42}}
}}
{{end-eqn}}
To show that there are no more such integers, we use the formulas:
{{begin-eqn}}
{{eqn | l = n
      | m = \prod_{p_i \mathop \divides n} {p_i}^{k_i}
      | ro = =
      | r = \prod_{j \mathop = 1}^r {p_i}^{k_i}
      | c = Prime Decomposition
}}
{{eqn | l = \map \phi n
      | m = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}
      | ro = =
      | r = \prod_{i \mathop = 1}^r p_i^{k_i} \paren {1 - \frac 1 p_i}
      | c = Euler Phi Function of Integer
}}
{{eqn | l = \map {\sigma_0} n
      | m = \prod_{i \mathop = 1}^r \paren {k_i + 1}
      | c = Divisor Count Function from Prime Decomposition
}}
{{eqn | l = \map {\sigma_1} n
      | m = \prod_{i \mathop = 1}^r \frac {p_i^{k_i + 1} - 1} {p_i - 1}
      | c = Divisor Sum of Integer
}}
{{end-eqn}}
Which arises from Euler Phi Function is Multiplicative, Divisor Count Function is Multiplicative and Divisor Sum Function is Multiplicative.
We define the mapping $f = \phi \cdot \sigma_0$ as follows:
:$\map f {p^k} = \map \phi {p^k} \map {\sigma_0} {p^k} = p^k \paren {1 - \dfrac 1 p} \paren {k + 1}$
where $p$ is a prime, and $k$ is a non-negative integer.
By Product of Multiplicative Functions is Multiplicative,  $f$ and $\dfrac f {\sigma_1}$ are multiplicative functions.
Then $\map \phi n \map {\sigma_0} n = \map {\sigma_1} n$ is equivalent to:
:$\ds \prod_{i \mathop = 1}^r \dfrac {\map f {p_i^{k_i} } } {\map {\sigma_1} {p_i^{k_i} } } = 1$
We consider the cases $k \ge 1$.
Then:
{{begin-eqn}}
{{eqn | l = \map f {p^k}
      | o = \ge
      | r = \map {\sigma_1} {p^k}
}}
{{eqn | ll= \leadsto
      | l = p^k \paren {1 - \frac 1 p} \paren {k + 1}
      | o = \ge
      | r = \frac {p^{k + 1} - 1} {p - 1}
}}
{{eqn | ll= \leadsto
      | l = p^{k - 1} \paren {p - 1}^2 \paren {k + 1}
      | o = \ge
      | r = p^{k + 1} - 1
}}
{{eqn | ll= \leadsto
      | l = k p^{k + 1} + \paren {k + 1} p^{k - 1} + 1
      | o = \ge
      | r = 2 \paren {k + 1} p^k
}}
{{end-eqn}}
Suppose equality holds.
If $k \ge 2$:
{{begin-eqn}}
{{eqn | l = k p^{k + 1} + \paren {k + 1} p^{k - 1} + 1
      | o = \ge
      | r = 2 \paren {k + 1} p^k
}}
{{eqn | ll = \leadsto
      | l = 1
      | o = \equiv
      | r = 0
      | rr = \pmod p
      | c = since $p^{k - 1}, p^k, p^{k + 1}$ are multiples of $p$
}}
{{end-eqn}} 
which is a contradiction.
This forces $k = 1$.
The equation becomes:
{{begin-eqn}}
{{eqn | l = p^2 + 2 + 1
      | r = 4 p
}}
{{eqn | ll = \leadsto
      | l = p^2 - 4 p + 3
      | r = 0
}}
{{eqn | ll = \leadsto
      | l = \paren {p - 1} \paren {p - 3}
      | r = 0
}}
{{end-eqn}}
Since $p \ne 1$, we must have $p = 3$.
Indeed:
:$\map f 3 = \map \phi 3 \map {\sigma_0} 3 = \map {\sigma_1} 3$
For the inequality case: $k p^{k + 1} + \paren {k + 1} p^{k - 1} + 1 > 2 \paren {k + 1} p^k$, we check each prime $p$ individually.
For $p = 2$:
{{begin-eqn}}
{{eqn | l = k \cdot 2^{k + 1} + \paren {k + 1} 2^{k - 1} + 1
      | o = >
      | r = 2 \paren {k + 1} 2^k
}}
{{eqn | ll= \leadsto
      | l = 4 k \paren {2^{k - 1} } + \paren {k + 1} \paren {2^{k - 1} }
      | o = \ge
      | r = 4 \paren {k + 1} \paren {2^{k - 1} }
      | c = as both sides are integers
}}
{{eqn | ll= \leadsto
      | l = 4 k + \paren {k + 1}
      | o = \ge
      | r = 4 \paren {k + 1}
}}
{{eqn | ll= \leadsto
      | l = k
      | o = \ge
      | r = 3
}}
{{end-eqn}}
For $p \ge 3$, we consider the criterion $k p \ge 2 \paren {k + 1}$.
This criterion implies $k p^{k + 1} \ge 2 \paren {k + 1} p^k$,
which in turn implies our inequality $k p^{k + 1} + \paren {k + 1} + 1 > 2 \paren {k + 1} p^k$.
For $p = 3$ and $k \ge 2$, $3 k = 2 k + k \ge 2 k + 2$.
For $p \ge 5$, $k p \ge 5 k > 2 k + 2 k \ge 2 k + 2$.
Therefore $\map f {p^k} > \map {\sigma_1} {p^k}$, except for:
:$\map f 3 = \map {\sigma_1} 3$
:$\map f 2 < \map {\sigma_1} 2$
:$\map f 4 < \map {\sigma_1} 4$
Now we try to find integers $n$ satisfying $\ds \prod_{i \mathop = 1}^r \dfrac {\map f {p_i^{k_i} } } {\map {\sigma_1} {p_i^{k_i} } } = 1$.
First observe that the sequence $\sequence {\dfrac {\map f {p^k} } {\map {\sigma_1} {p^k} } }_{k \mathop = 1}^\infty$ is strictly increasing:
{{begin-eqn}}
{{eqn | l = \frac {\map f {p^{k + 1} } } {\map {\sigma_1} {p^{k + 1} } } - \frac {\map f {p^k} } {\map {\sigma_1} {p^k} }
      | r = \frac {p^k \paren {1 - 1 / p} \paren {k + 2} } {\paren {p^{k + 2} - 1} / \paren {p - 1} } - \frac {p^{k - 1} \paren {1 - 1 / p} \paren {k + 1} } {\paren {p^{k + 1} - 1} / \paren {p - 1} }
}}
{{eqn | r = \frac {p^{k - 1} \paren {p - 1}^2 \paren {k + 2} } {p^{k + 2} - 1} - \frac {p^{k - 2} \paren {p - 1}^2 \paren {k + 1} } {p^{k + 1} - 1}
}}
{{eqn | r = \frac {p^{k - 2} \paren {p - 1}^2} {\paren {p^{k + 2} - 1} \paren {p^{k + 1} - 1} } \paren {p \paren {p^{k + 1} - 1} \paren {k + 2} - \paren {p^{k + 2} - 1} \paren {k + 1} }
}}
{{eqn | r = \frac {p^{k - 2} \paren {p - 1}^2} {\paren {p^{k + 2} - 1} \paren {p^{k + 1} - 1} } \paren {p^{k + 2} - \paren {k + 2} p + k + 1}
}}
{{eqn | r = \frac {p^{k - 2} \paren {p - 1}^2} {\paren {p^{k + 2} - 1} \paren {p^{k + 1} - 1} } \paren {1 + \paren {k + 2} \paren {p - 1} + \sum_{m \mathop = 2}^{k + 2} \dbinom {k + 2} m \paren {p - 1}^m - \paren {k + 2} p + k + 1} 
      | c = Binomial Theorem
}}
{{eqn | r = \frac {p^{k - 2} \paren {p - 1}^2} {\paren {p^{k + 2} - 1} \paren {p^{k + 1} - 1} } \sum_{m \mathop = 2}^{k + 2} \dbinom {k + 2} m \paren {p - 1}^m
}}
{{eqn | o = >
      | r = 0
      | c = as the sum is non-empty and each term is strictly positive
}}
{{end-eqn}}
Now we consider these particular values for $\dfrac f {\sigma_1}$:
:$\dfrac {\map f 2} {\map {\sigma_1} 2 } = \dfrac 2 3$
:$\dfrac {\map f 4} {\map {\sigma_1} 4} = \dfrac 6 7$
:$\dfrac {\map f 3} {\map {\sigma_1} 3} = 1$
:$\dfrac {\map f 5} {\map {\sigma_1} 5} = \dfrac 4 3$
:$\dfrac {\map f {25} } {\map {\sigma_1} {25} } = \dfrac {60} {31} > \dfrac 3 2$
:$\dfrac {\map f 7} {\map {\sigma_1} 7} = \dfrac 3 2$
And for $p \ge 11$:
{{begin-eqn}}
{{eqn | l = \frac {\map f p} {\map {\sigma_1} p}
      | r = \frac {2 \paren {p - 1} } {\paren {p^2 - 1} / \paren {p - 1} }
}}
{{eqn | r = \frac {2 \paren {p - 1} } {p + 1}
}}
{{eqn | r = 2 - \frac 4 {p + 1}
}}
{{eqn | o = \ge
      | r = 2 - \frac 4 {12} > \frac 3 2
}}
{{end-eqn}}
By the strictly increasing property above, every higher power of these primes has $\dfrac f {\sigma_1} > \dfrac 3 2$.
Since $\dfrac f {\sigma_1}$ is multiplicative, only $n = 3, 14, 42$ can give $\dfrac {\map f n} {\map {\sigma_1} n} = 1$.
Since $\map \phi 1 \map {\sigma_0} 1 = \map {\sigma_1} 1$, the only positive integers whose Euler $\phi$ function multiplied by its  divisor count function equals its divisor sum are $1, 3, 14, 42$.
{{qed}}