Numbers whose Fourth Root equals Number of Divisors
Theorem
There are $4$ positive integers whose $4$th root equals the number of its divisors:
\(\ds 1\) | \(=\) | \(\ds 1^4\) | ||||||||||||
\(\ds 625\) | \(=\) | \(\ds 5^4\) | ||||||||||||
\(\ds 6561\) | \(=\) | \(\ds 9^4\) | ||||||||||||
\(\ds 4 \, 100 \, 625\) | \(=\) | \(\ds 45^4\) |
This sequence is A143026 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
Proof
\(\ds \map {\sigma_0} 1\) | \(=\) | \(\ds 1\) | $\sigma_0$ of $1$ | |||||||||||
\(\ds \map {\sigma_0} {625}\) | \(=\) | \(\ds 5\) | $\sigma_0$ of $625$ | |||||||||||
\(\ds \map {\sigma_0} {6561}\) | \(=\) | \(\ds 9\) | $\sigma_0$ of $6561$ | |||||||||||
\(\ds \map {\sigma_0} {4 \, 100 \, 625}\) | \(=\) | \(\ds 45\) | $\sigma_0$ of $4 \, 100 \, 625$ |
Suppose $N = \map {\sigma_0} {N^4}$.
By Divisor Count Function is Odd Iff Argument is Square, $N$ must be odd.
The case $N = 1$ is trivial.
Suppose $N$ is a prime power.
Write $N = p^n$.
By Divisor Count Function of Power of Prime:
- $N = \map {\sigma_0} {p^{4 n} } = 4 n + 1$
- $N = p^n \ge 1 + n \paren {p - 1}$
This gives us the inequality:
- $4 n + 1 \ge 1 + n \paren {p - 1}$
which can be simplified to:
- $4 \ge p - 1$
The only odd primes satisfying the inequality are $3$ and $5$.
We have:
- $\map {\sigma_0} {3^4} = 5 > 3^1$
- $\map {\sigma_0} {3^8} = 9 = 3^2$
- $\map {\sigma_0} {3^{4 n} } = 4 n + 1 < 3^n$ for $n > 2$
- $\map {\sigma_0} {5^4} = 5 = 5^1$
- $\map {\sigma_0} {5^{4 n} } = 4 n + 1 < 5^n$ for $n > 1$
- $\map {\sigma_0} {p^{4 n} } = 4 n + 1 < p^n$ for any $p > 5$
Hence $625$ and $6561$ are the only prime powers satisfying the property.
Note that Divisor Count Function is Multiplicative.
To form an integer $N$ with our property, we must choose and multiply prime powers from the list above.
The product $625 \times 6561 = 4 \, 100 \, 625$ gives equality.
If we chose any $\tuple {p, n}$ with $\map {\sigma_0} {p^{4 n} } < p^n$, we must choose $3^4$ in order for equality to possibly hold.
Then $\map {\sigma_0} {3^4} = 5 \divides N$, so a tuple $\tuple {5, n}$ must be chosen.
If $\tuple {5, 1}$ was chosen, $5^2 \nmid N$.
But $\map {\sigma_0} {3^4 \times 5^4} = 25 \divides N$, which is a contradiction.
Suppose $\tuple {5, n}$ with $n \ge 2$ was chosen.
Then $\map {\sigma_0} {3^4 \times 5^{4 n} } = 20 n + 1 < 3 \times 5^n$, a contradiction.
Thus we have exhausted all cases.
$\blacksquare$
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $4,100,625$