Prime whose Divisor Sum is Square is 3

Theorem

There is exactly $1$ prime number whose divisor sum is a square number, and that is $3$:

$\map {\sigma_1} 3 = 4$

Proof

That $\map {\sigma_1} 3 = 4$ is shown at $\sigma_1$ of $3$.

It remains to be shown there are no more.

Let $n \in \N$ such that $\map {\sigma_1} n$ is square.

$\map {\sigma_1} n = m^2$

Suppose $n$ is prime.

From Divisor Sum of Prime Number:

$\map {\sigma_1} n = n + 1$

So we have:

\(\ds n + 1\)

\(=\)

\(\ds m^2\)

\(\ds \leadsto \ \ \)

\(\ds n\)

\(=\)

\(\ds m^2 - 1\)

\(\ds \)

\(=\)

\(\ds \paren {m + 1} \paren {m - 1}\)

Difference of Two Squares

But $\paren {m + 1} \paren {m - 1}$ is prime if and only if $m - 1 = 1$, that is, $m = 2$.

That is the case covered by $\map {\sigma_1} 3 = 4$.

Hence the result.

$\blacksquare$