Integers whose Divisor Count equals Cube Root

Theorem
There are $3$ positive integers whose divisor counting function equals its cube root:

Proof
Suppose $N = \map {\sigma_0} {N^3}$.

The case $N = 1$ is trivial.

Suppose $N$ is a prime power.

Write $N = p^n$.

By Divisor Counting Function of Power of Prime:
 * $N = \map {\sigma_0} {p^{3 n} } = 3 n + 1$

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

This gives us the inequality:


 * $3 n + 1 \ge 1 + n \paren {p - 1}$

which can be simplified to:


 * $3 \ge p - 1$

The only primes satisfying the inequality are $2$ and $3$.

We have:


 * $\map {\sigma_0} {2^3} = 4 > 2^1$


 * $\map {\sigma_0} {2^6} = 7 > 2^2$


 * $\map {\sigma_0} {2^9} = 10 > 2^3$


 * $\map {\sigma_0} {2^{3 n} } = 3 n + 1 < 2^n$ for $n > 3$


 * $\map {\sigma_0} {3^3} = 4 > 3^1$


 * $\map {\sigma_0} {3^{3 n} } = 3 n + 1 < 3^n$ for $n > 1$


 * $\map {\sigma_0} {p^{3 n} } = 3 n + 1 < p^n$ for all $p > 3$

Hence no prime powers satisfy 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.

If we chose any $\tuple {p, n}$ with $\map {\sigma_0} {p^{4 n} } < p^n$, we must choose $2^m$ or $3^1$ in order for equality to possibly hold.

If $\tuple {2, 1}$ was chosen, $2^2 \nmid N$.

But $\map {\sigma_0} {2^3} = 4 \divides N$, which is a contradiction.

Suppose $\tuple {2, 2}$ was chosen.

Then $\map {\sigma_0} {2^6} = 7 \divides N$.

Then we must choose some $\tuple {7, n}$.

For $n = 1$, $\map {\sigma_0} {7^3} = 4$.


 * $\map {\sigma_0} {2^6 \times 7^3} = 4 \times 7 = 28 = 2^2 \times 3$


 * $\map {\sigma_0} {2^6 \times 7^3 \times p^{3 m} } = 28 \paren {3 m + 1} < 28 \times p^m$ for all $p \ne 2, 7$

For $n > 1$, $\map {\sigma_0} {2^6 \times 7^{3 n} } = 7 \paren {3 n + 1} < 4 \times 7^n$, a contradiction.

Suppose $\tuple {2, 3}$ was chosen.

Then:
 * $\map {\sigma_0} {2^9} = 10 \divides N$

Then we must choose some $\tuple {5, n}$.

For $n = 1$, $\map {\sigma_0} {5^3} = 4$.


 * $\map {\sigma_0} {2^9 \times 5^3} = 10 \times 4 = 40 = 2^3 \times 5$


 * $\map {\sigma_0} {2^9 \times 5^3 \times p^{3 m} } = 40 \paren {3 m + 1} < 40 \times p^m$ for all $p \ne 2, 5$

For $n > 1$, $\map {\sigma_0} {2^9 \times 5^{3 n} } = 10 \paren {3 n + 1} < 2^3 \times 5^n$, a contradiction.

Suppose $\tuple {3, 1}$ was chosen.

Then $\map {\sigma_0} {3^3} = 4 \divides N$.

Then this case coincides the cases above.

Thus we have exhausted all cases.