Is that you, Mersenne?

Let $a=256=2^8$ . Then

$\begin{aligned} f(n) & = f(255) f(256) = f(a-1)f(a) \\ n^2 + n + 1 & = \left((a-1)^2+a-1+1\right)\left(a^2+a+1\right) \\ & = \left(a^2-a+1\right)\left(a^2+a+1\right) \\ & = a^4 + a^2 + 1 \\ \implies n & = a^2 = 2^{16} \end{aligned}$

Then the sum of the divisors of $n=2^{16}$ is given by $\displaystyle \sum_{k=0}^{16} 2^k = \dfrac {2^{17}-1}{2-1} = \boxed {131071}$ .

Notice that $f(x-1) = (x-1)^2 + (x-1) + 1 = (x^2 - 2x + 1) + (x-1) + 1 = x^2 - x + 1.$

Then the key is to observe that $f(x-1)f(x) = (x^2 - x + 1)(x^2 + x + 1) = (x^2 + 1)^2 - x^2 = \left(x^2\right)^2 + x^2 + 1 = f(x^2),$ namely, ${\color{#D61F06} f(x-1)f(x) = f(x^2)},$ for all $x$ . (P.S., it’s more fun to derive this identity by studying the sixth roots of unity , and notice that both $f(x)$ and $f(x-1)$ are cyclotomic polynomials .)

In particular, by taking $x = 256$ , we have $f(255)f(256) = f(256^2) = f(2^{16}) \overset{\text{set}}{=} f(n),$ which means $n = 2^{16}$ as the function $f: \mathbb{R} \longrightarrow \mathbb{R}$ defined by $f(x) = x^2 + x + 1$ is strictly increasing for $x \ge -\frac{1}{2}$ , namely, $f$ is one-to-one on $[-\frac{1}{2}, \infty)$ .

Hence, the sum of all positive divisors of $n$ is $\sigma(n) = \underbrace{2^0 + 2^1 + 2^2 + \cdots + 2^{16}}_{\text{geometric series}} = 2^{17} - 1 = \boxed{\mathbf{131071}},$ which is in fact a Mersenne prime . (Here $\sigma(n)$ denotes the sum of all positive divisors of $n$ .)