Integers whose Phi times Divisor Count equal Divisor Sum
Theorem
The positive integers whose Euler $\phi$ function multiplied by its divisor count 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 Count 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 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:
\(\ds \map f {p^k}\) | \(\ge\) | \(\ds \map {\sigma_1} {p^k}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds p^k \paren {1 - \frac 1 p} \paren {k + 1}\) | \(\ge\) | \(\ds \frac {p^{k + 1} - 1} {p - 1}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds p^{k - 1} \paren {p - 1}^2 \paren {k + 1}\) | \(\ge\) | \(\ds p^{k + 1} - 1\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds k p^{k + 1} + \paren {k + 1} p^{k - 1} + 1\) | \(\ge\) | \(\ds 2 \paren {k + 1} p^k\) |
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 count function equals its divisor sum are $1, 3, 14, 42$.
$\blacksquare$
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $42$