In the 1986 book *An Introduction to the Theory of Surreal Numbers*, Gonshor, on page 117, notes that it is an open problem whether $\omega(\sqrt{2}+1)+1$ is a prime, using the standard definition of integral number he introduces earlier in the chapter.

I'll give a little background, as requested, using Gonshor as a reference.

The surreal numbers are a superclass of the infinitesimals, though I'm not familiar with their exact sizes.

A surreal number is a function from an initial segment of the ordinals into $\{+,-\}$. (An ordinal sequence which terminates)

Then, given two surreals, $a$, $b$, let $a < b$ if $a(\alpha) < b(\alpha)$ where $\alpha$ is the first place they differ, with the convention $- < 0 < +$, where $0$ is an abuse of notation that simply means that the function is undefined at that point. So, $(++) > (+) > (+-)$. Some investigation gives that $0\sim (),\ 1 \sim (+),\ 2 \sim (++), \ldots$ Most of the book is spent on embedding the reals and infinitesimals in the surreals.

Then you can define the generalized integers in the surreals: $a$ is an integer if the exponents in normal form of $a$ are non-negative, and if a $0$ exponent occurs, then the real coefficient is an integer. Normal form is a bit complicated to define, but basically you write numbers in terms of $\omega$ and $\epsilon$ where $\omega$ is the first infinite ordinal, so $(+++...)$ times and $\epsilon = 1/\omega$, which also exists in the surreals. So, $1/3\omega^2+3$ is a generalized integer, and so is $\sqrt{\omega}+2$, but not $0.5+\omega^{-1}$, as both terms violate the definition.

For the definition of prime, $1=(+)$ is still a unit, so primality amounts to proving that a surreal has more than two factors (itself and $1$). $\omega$ is factorizable since $n \in \mathbb{N}$ and $\omega/n$ are both generalized integers.

That trick doesn't work with $\omega(\sqrt{2}+1)+1$.

Is this still an open problem? The book is a couple decades old (picked it up at the library), and searching for surreal numbers doesn't turn up many results.

Thanks