Is the surreal number $\omega(\sqrt{2}+1)+1$ a prime? 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
 A: This is not an answer but it is too long for a comment. I think that the question is still open, and that in general there are no known prime elements with finite support besides primes in $\mathbb{Z}$.

There are a few papers, besides Conway and Gonshor's books, that deal with this type of question. A more general problem is the existence of prime or irreducibles in integer parts of real-closed fields, such as $\mathbf{Oz}$ (the "omnific integers"), which can be written as a Hahn series ring $K((G^{\leq 0}))$ where $G$ is an additively denoted ordered group.
For those with more emphasis on $\mathbf{Oz}$, you can look at Alessandro Berarducci's Factorization in generalized power series (Trans. Amer. Math. Soc. 352 (2000) 553-577, doi:10.1090/S0002-9947-99-02172-8), which I think first found out that one could apply valuation theoretic insight to this question. A more recent paper which in particular gives more recent references for this type of question is Sonia L'Innocente and Vincenzo Mantova's preprint Factorisation theorems for generalised power series, arXiv:1710.07304.
Berarducci gave a positive answer to Conway's conjecture that $\omega+\omega^{\frac{1}{2}}+\omega^{\frac{1}{3}}+ \cdot \cdot \cdot+1$ is irreducible. Daniel Pitteloud subsequently proved that this number is prime in $\mathbf{Oz}$.
L'Innocente and Mantova generalised some results and proved that certain numbers with infinite support as series are prime.
A: The answer is no. The factorization is $(\sqrt{\sqrt{2}+1}\omega^{1/2} - \sqrt{2\sqrt{\sqrt{2}+1}}\omega^{1/4}+1)(\sqrt{\sqrt{2}+1}\omega^{1/2} + \sqrt{2\sqrt{\sqrt{2}+1}}\omega^{1/4}+1)$
