Integers whose Phi times Divisor Count equal Divisor Sum

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Theorem

The positive integers whose Euler $\phi$ function multiplied by its divisor counting function equals its divisor sum are:

$1, 3, 14, 42$

This sequence is A104905 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof

\(\ds \map \phi 1 \map {\sigma_0} 1\) \(=\) \(\ds 1 \times 1\) $\phi$ of $1$, $\sigma_0$ of $1$
\(\ds \) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds \map {\sigma_1} 1\) $\sigma_1$ of $1$


\(\ds \map \phi 3 \map {\sigma_0} 3\) \(=\) \(\ds 2 \times 2\) $\phi$ of $3$, $\sigma_0$ of $3$
\(\ds \) \(=\) \(\ds 4\)
\(\ds \) \(=\) \(\ds \map {\sigma_1} 3\) $\sigma_1$ of $3$


\(\ds \map \phi {14} \map {\sigma_0} {14}\) \(=\) \(\ds 6 \times 4\) $\phi$ of $14$, $\sigma_0$ of $14$
\(\ds \) \(=\) \(\ds 24\)
\(\ds \) \(=\) \(\ds \map {\sigma_1} {14}\) $\sigma_1$ of $14$


\(\ds \map \phi {42} \map {\sigma_0} {42}\) \(=\) \(\ds 12 \times 8\) $\phi$ of $42$, $\sigma_0$ of $42$
\(\ds \) \(=\) \(\ds 96\)
\(\ds \) \(=\) \(\ds \map {\sigma_1} {42}\) $\sigma_1$ of $42$


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

\(\ds n\) \(=\) \(\, \ds \prod_{p_i \mathop \divides n} {p_i}^{k_i} \, \) \(\, \ds = \, \) \(\ds \prod_{j \mathop = 1}^r {p_i}^{k_i}\) Prime Decomposition
\(\ds \map \phi n\) \(=\) \(\, \ds n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p} \, \) \(\, \ds = \, \) \(\ds \prod_{i \mathop = 1}^r p_i^{k_i} \paren {1 - \frac 1 p_i}\) Euler Phi Function of Integer
\(\ds \map {\sigma_0} n\) \(=\) \(\, \ds \prod_{i \mathop = 1}^r \paren {k_i + 1} \, \) \(\ds \) Divisor Counting Function from Prime Decomposition
\(\ds \map {\sigma_1} n\) \(=\) \(\, \ds \prod_{i \mathop = 1}^r \frac {p_i^{k_i + 1} - 1} {p_i - 1} \, \) \(\ds \) Divisor Sum of Integer

Which arises from Euler Phi Function is Multiplicative, Divisor Counting Function is Multiplicative and Divisor Sum 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 $\dfrac f {\sigma_1}$ are multiplicative functions, by Product of Multiplicative Functions is Multiplicative.

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:

\(\ds \map f {p^k}\) \(\ge\) \(\ds \map {\sigma_1} {p^k}\)
\(\ds \leadsto \ \ \) \(\ds p^k \paren {1 - \frac 1 p} \paren {k + 1}\) \(\ge\) \(\ds \frac {p^{k + 1} - 1} {p - 1}\)
\(\ds \leadsto \ \ \) \(\ds p^{k - 1} \paren {p - 1}^2 \paren {k + 1}\) \(\ge\) \(\ds p^{k + 1} - 1\)
\(\ds \leadsto \ \ \) \(\ds k p^{k + 1} + \paren {k + 1} p^{k - 1} + 1\) \(\ge\) \(\ds 2 \paren {k + 1} p^k\)

Suppose equality holds.

If $k \ge 2$,

\(\ds k p^{k + 1} + \paren {k + 1} p^{k - 1} + 1\) \(\ge\) \(\ds 2 \paren {k + 1} p^k\)
\(\ds \leadsto \ \ \) \(\ds 1\) \(\equiv\) \(\ds 0\) \(\ds \pmod p\) since $p^{k - 1}, p^k, p^{k + 1}$ are multiples of $p$

Which is a contradiction.

This forces $k = 1$.


The equation becomes:

\(\ds p^2 + 2 + 1\) \(=\) \(\ds 4 p\)
\(\ds \leadsto \ \ \) \(\ds p^2 - 4 p + 3\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \paren {p - 1} \paren {p - 3}\) \(=\) \(\ds 0\)

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 $p = 2$:

\(\ds k \cdot 2^{k + 1} + \paren {k + 1} 2^{k - 1} + 1\) \(>\) \(\ds 2 \paren {k + 1} 2^k\)
\(\ds \leadsto \ \ \) \(\ds 4 k \paren {2^{k - 1} } + \paren {k + 1} \paren {2^{k - 1} }\) \(\ge\) \(\ds 4 \paren {k + 1} \paren {2^{k - 1} }\) as both sides are integers
\(\ds \leadsto \ \ \) \(\ds 4 k + \paren {k + 1}\) \(\ge\) \(\ds 4 \paren {k + 1}\)
\(\ds \leadsto \ \ \) \(\ds k\) \(\ge\) \(\ds 3\)


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:

\(\ds \frac {\map f {p^{k + 1} } } {\map {\sigma_1} {p^{k + 1} } } - \frac {\map f {p^k} } {\map {\sigma_1} {p^k} }\) \(=\) \(\ds \frac {p^k \paren {1 - 1 / p} \paren {k + 2} } {\paren {p^{k + 2} - 1} / \paren {p - 1} } - \frac {p^{k - 1} \paren {1 - 1 / p} \paren {k + 1} } {\paren {p^{k + 1} - 1} / \paren {p - 1} }\)
\(\ds \) \(=\) \(\ds \frac {p^{k - 1} \paren {p - 1}^2 \paren {k + 2} } {p^{k + 2} - 1} - \frac {p^{k - 2} \paren {p - 1}^2 \paren {k + 1} } {p^{k + 1} - 1}\)
\(\ds \) \(=\) \(\ds \frac {p^{k - 2} \paren {p - 1}^2} {\paren {p^{k + 2} - 1} \paren {p^{k + 1} - 1} } \paren {p \paren {p^{k + 1} - 1} \paren {k + 2} - \paren {p^{k + 2} - 1} \paren {k + 1} }\)
\(\ds \) \(=\) \(\ds \frac {p^{k - 2} \paren {p - 1}^2} {\paren {p^{k + 2} - 1} \paren {p^{k + 1} - 1} } \paren {p^{k + 2} - \paren {k + 2} p + k + 1}\)
\(\ds \) \(=\) \(\ds \frac {p^{k - 2} \paren {p - 1}^2} {\paren {p^{k + 2} - 1} \paren {p^{k + 1} - 1} } \paren {1 + \paren {k + 2} \paren {p - 1} + \sum_{m \mathop = 2}^{k + 2} \dbinom {k + 2} m \paren {p - 1}^m - \paren {k + 2} p + k + 1}\) Binomial Theorem
\(\ds \) \(=\) \(\ds \frac {p^{k - 2} \paren {p - 1}^2} {\paren {p^{k + 2} - 1} \paren {p^{k + 1} - 1} } \sum_{m \mathop = 2}^{k + 2} \dbinom {k + 2} m \paren {p - 1}^m\)
\(\ds \) \(>\) \(\ds 0\) as the sum is non-empty and each term is strictly positive


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 $p \ge 11$:

\(\ds \frac {\map f p} {\map {\sigma_1} p}\) \(=\) \(\ds \frac {2 \paren {p - 1} } {\paren {p^2 - 1} / \paren {p - 1} }\)
\(\ds \) \(=\) \(\ds \frac {2 \paren {p - 1} } {p + 1}\)
\(\ds \) \(=\) \(\ds 2 - \frac 4 {p + 1}\)
\(\ds \) \(\ge\) \(\ds 2 - \frac 4 {12} > \frac 3 2\)

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 counting function equals its divisor sum are $1, 3, 14, 42$.

$\blacksquare$


Sources