Integers whose Phi times Divisor Count equal Divisor Sum

Theorem
The positive integers whose Euler $\phi$ function multiplied by its divisor counting function equals its $\sigma$ function are:
 * $1, 3, 14, 42$

Proof
To show that there are no more such integers, we use the formulas:

Which arises from Euler Phi Function is Multiplicative, Divisor Counting Function is Multiplicative and Sigma Function is Multiplicative.

We define the function $f = \phi \cdot \sigma_0$ 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.

Note that $f$ and $\dfrac f \sigma$ are multiplicative functions, by Product of Multiplicative Functions is Multiplicative.

Then $\map \phi n \map {\sigma_0} n = \map \sigma n$ is equivalent to:


 * $\ds \prod_{i \mathop = 1}^r \dfrac {\map f {p_i^{k_i} } } {\map \sigma {p_i^{k_i} } } = 1$

We consider the cases $k \ge 1$. Then:

Suppose equality holds.

If $k \ge 2$,

Which is a contradiction.

This forces $k = 1$.

The equation becomes:

Since $p \ne 1$, we must have $p = 3$.

Indeed:
 * $\map f 3 = \map \phi 3 \map {\sigma_0} 3 = \map \sigma 3$

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 $p = 2$:

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 $\map f {p^k} > \map \sigma {p^k}$, except for:
 * $\map f 3 = \map \sigma 3$
 * $\map f 2 < \map \sigma 2$
 * $\map f 4 < \map \sigma 4$

Now we try to find integers $n$ satisfying $\ds \prod_{i \mathop = 1}^r \dfrac {\map f {p_i^{k_i} } } {\map \sigma {p_i^{k_i} } } = 1$.

First observe that the sequence $\sequence {\dfrac {\map f {p^k} } {\map \sigma {p^k} } }_{k \mathop = 1}^\infty$ is strictly increasing:

Now we consider these particular values for $\dfrac f \sigma$:


 * $\dfrac {\map f 2 } {\map \sigma 2 } = \dfrac 2 3$
 * $\dfrac {\map f 4 } {\map \sigma 4 } = \dfrac 6 7$
 * $\dfrac {\map f 3 } {\map \sigma 3 } = 1$
 * $\dfrac {\map f 5 } {\map \sigma 5 } = \dfrac 4 3$
 * $\dfrac {\map f {25} } {\map \sigma {25} } = \dfrac {60} {31} > \dfrac 3 2$
 * $\dfrac {\map f 7 } {\map \sigma 7 } = \dfrac 3 2$

And for $p \ge 11$:

By the strictly increasing property above, every higher power of these primes has $\dfrac f \sigma > \dfrac 3 2$.

Since $\dfrac f \sigma$ is multiplicative, only $n = 3, 14, 42$ can give $\dfrac {\map f n} {\map \sigma n} = 1$.

Since $\map \phi 1 \map {\sigma_0} 1 = \map \sigma 1$, the only positive integers whose Euler $\phi$ function multiplied by its divisor counting function equals its $\sigma$ function are $1, 3, 14, 42$.