Consider the numbers of the form: $$A\_n = \prod\_{\pm}\left(\pm 1\pm \sqrt{2} \pm \cdots \pm \sqrt{n}\right)$$


where, the product in taken oven all $2^n$ terms with variations in sign. We know such numbers are integers (even perfect-squares). 



> 
> Is there a way to investigate the asymptotics of the largest prime divisor of $A\_n$?
> 
> 
> 


Source: [Problem A-578](http://www.komal.hu/verseny/feladat.cgi?a=honap&h=201301&t=mat&l=en) where it was claimed that that for all $\varepsilon>0$ there exists an $n\_0$ such that for every $n>n\_0$ the largest prime divisor of $A\_n$ is smaller than $\displaystyle 2^{2^{\,\huge{\varepsilon} n}}$. I haven't been able to prove this claim either.



Estimating size of greatest prime divisor of a sequence of integers
The sequence grows too rapidly and I doubt that one can get meaningful asymptotics for the largest prime factor. However, one can show that the largest prime factor is bounded by $(n\sqrt{n})^{2^{\pi(n)}}$ where $\pi(n)$ is the number of primes up to $n$, and since $\pi(n) \sim n/\log n$, this implies a stronger bound than what you wanted. 


To see this, put $\alpha =1+ \sqrt{2} + \ldots + \sqrt{n}$ and let $\epsilon=(\epsilon\_1,\ldots,\epsilon\_n)$ be any choice of signs. Set $\epsilon(\alpha)= \sum\_{j=1}^{n} \epsilon\_j \sqrt{j}$. Note that $\alpha$ and all the $2^n$ values $\epsilon(\alpha)$ all belong to the field $K={\Bbb Q}(\sqrt{p}, 2\le p\le n)$ which has degree $2^{\pi(n)}$ and Galois group $H \simeq ({\Bbb Z}/2)^{\pi(n)}$. Note that the choice of all possible signs $\epsilon$ forms a group $G \simeq ({\Bbb Z}/2)^n$, and $H$ may be thought of as a subgroup of $G$ in the obvious way. Then 
$$ 
A\_n = \prod\_{\epsilon \in G/H} N\_K(\epsilon(\alpha)), 
$$ 
where the product is over the $2^{n-\pi(n)}$ cosets $G/H$, and $N\_K$ is the norm in $K$ (the product over all the elements of $H$). Note that all the $N\_K(\epsilon(\alpha))$ are non-zero integers. Further all these norms are bounded by $(n\sqrt{n})^{2^{\pi(n)}}$ (there are $2^{\pi(n)}$ conjugates and each is bounded in size by $\sqrt{1} +\ldots + \sqrt{n} \le n\sqrt{n}$). 


Thus $A\_n$ can be written as the product of $2^{n-\pi(n)}$ integers each of size at most $(n\sqrt{n})^{2^{\pi(n)}}$, and therefore its largest prime factor is no more than that. 


