IMO! 2

Let $y=n^2+1$ ; that means that $n=\sqrt{y-1}$ . This allows us to rewrite $2n+\sqrt{2n}$ as $2\sqrt{y-1}+\sqrt{2\sqrt{y-1}}$ . Now, we can rewrite the problem as the search of the values of $y$ for which exists a prime divisor greater than $2\sqrt{y-1}+\sqrt{2\sqrt{y-1}}$ (that can be done because $n$ is a positive integer and so $y=n^2+1$ is a biective function in that interval: that means, for each value of $n$ there will be one and only one corresponding value of $y$ and vice versa). To solve this problem, first consider the fact that the more prime factors a number has, the smaller they will necessarily be; we need a prime factor greater than a value, so it would be nice to have a factor as big as possible: that means we should consider non-prime numbers formed by the lowest possible number of factors, wich is 2. Now, we need that one of the two factors is the biggest possible (it will be the factor greater than $2\sqrt{y-1}+\sqrt{2\sqrt{y-1}}$ ); to do this, we have to minimize the other factor: the lowest value it can have is 2. All that leads us to consider numbers of the form $2p$ , where $p$ is a prime; $p$ can be written as $\frac{y}{2}$ . Now, let's consider the disequation $y/2>2\sqrt{y-1}+\sqrt{2\sqrt{y-1}}$ : without solving it, we can notice that the left member has degree 1, while the second has degree $\frac{1}{2}$ : that means that there is a value $k$ after that the first member is greater than the second one, and so there is an intervall of the form $]k,\infty[$ in which the disequation is verified (this is true only because the numbers involved are positive). There are infinite primes contained in that interval, hence there are infinite primes $p=\frac{y}{2}$ that are greater than $2\sqrt{y-1}+\sqrt{2\sqrt{y-1}}$ . From that, we can assume that there are infinite values of $y$ (and so of $n$ ) that verify our initial condition.