2008 IMO Problems/Problem 3
The main idea is to take a gaussian prime and multiply it by a "twice as small" to get . The rest is just making up the little details.
Solution
For each sufficiently large prime of the form , we shall find a corresponding satisfying the required condition with the prime number in question being . Since there exist infinitely many such primes and, for each of them, , we will have found infinitely many distinct satisfying the problem.
Take a prime of the form and consider its "sum-of-two squares" representation , which we know to exist for all such primes. As , assume without loss of generality that . If , then is our guy, and as long as (and hence ) is large enough. Let's see what happens when .
Since and are (obviously) co-prime, there must exist integers and such that In fact, if and are such numbers, then and work as well for any integer , so we can assume that .
Define and let's see why this was a good choice. For starters, notice that .
If , then from (1), we see that must divide and hence . In turn, and . Therefore, and so , from where . Finally, and the case is cleared.
We can safely assume now that As implies , we have so
Before we proceed, we would like to show that . Observe that the function over reaches its minima on the ends, so given is minimized for , where it equals . So we want to show that which obviously holds for large .
Now armed with and (2), we get where
Finally,