2008 IMO Problems/Problem 3

(still editing...)

For each sufficiently large prime $p$ of the form $4k+1$ , we shall find a corresponding $n$ satisfying the required condition with the prime number in question being $p$ . Since there exist infinitely many such primes and, for each of them, $n \ge \sqrt{p-1}$ , we will have found infinitely many distinct $n$ satisfying the problem.

Take a prime $p$ of the form $4k+1$ and consider its "sum-of-two squares" representation $p=a^2+b^2$ , which we know to exist for all such primes. If $a=1$ or $b=1$ , then $n=b$ or $n=a$ is our guy, and $p=n^2+1 > 2n+\sqrt(2n)$ as long as $p$ (and hence $n$ ) is large enough.

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2008 IMO Problems/Problem 3