Numbers whose Fourth Root equals Number of Divisors

Theorem
There are $4$ positive integers whose $4$th root equals the number of its divisors:

Proof
Suppose $N = \map {\sigma_0} {N^4}$.

By Divisor Counting 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 Counting Function of Power of Prime:
 * $N = \map {\sigma_0} {p^{4 n} } = 4 n + 1$

By Bernoulli's Inequality:
 * $N = p^n \ge 1 + n \paren {p - 1}$

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 Counting 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.