2008 IMO Problems/Problem 3

(still editing...)

The main idea is to take a gaussian prime $a+bi$ and multiply it by a "twice smaller" $c+di$ to get $n+i$ . The rest is just making up the little details.

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. As $a\ne b$ , assume without loss of generality that $b>a$ . If $a=1$ , then $n=b$ is our guy, and $p=n^2+1 > 2n+\sqrt{2n}$ as long as $p$ (and hence $n$ ) is large enough. Let's see what happens when $b>a>1$ .

Since $a$ and $b$ are (obviously) co-prime, there must exist integers $c$ and $d$ such that $ad+bc=1 \leqno{(1)}$ In fact, if $c$ and $d$ are such numbers, then $c\pm a$ and $d\mp b$ work as well, so we can assume that $c \in \left[-\frac{a}{2}, \frac{a}{2}\right]$ .

Define $n=|ac-bd|$ and let's see what happens. Notice that $(a^2+b^2)(c^2+d^2)=n^2+1$ .

If $c=\pm\frac{a}{2}$ , then from (1), we get $a\2$ (Error compiling LaTeX. Unknown error_msg) and hence $a=2$ . That means that $\displaystyle d=-\frac{b-1}{2}$ and $n=\frac{b(b-1)}{2}-2$ . Therefore, $b^2-b=2n+4>2n$ and so $\left(b-\frac{1}{2}\right)^2>2n$ , from where $b > \sqrt{2n}+\frac{1}{2}$ . Finally, $p=b^2+2^2 > 2n+\sqrt{2n}$ and the case $c=\pm\frac{a}{2}$ is cleared.

We can safely assume now that $|c| \le \frac{a-1}{2}.$ Automatically, $|d| \le \frac{b-1}{2}$ , since $2|d| = 2|\frac{1-bc}{a} | \le \frac{b(a-1)+2}{a} < \frac{ba}{a} = b,$ as $b>a>1$ implies $b>2$ . We have $n^2+1 = (a^2+b^2)(c^2+d^2) > p \left( \frac{(a-1)^2}{4}+\frac{(b-1)^2}{4} \right).$

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2008 IMO Problems/Problem 3