2008 IMO Problems/Problem 3

Revision as of 21:29, 3 September 2008 by Vbarzov (talk | contribs)

(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. 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. Let's see what happens when both $a>1$ and $b>1$. Apparently, $a\ne b$, so assume without loss of generality that $b>a>1$.

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 $d=-\frac{b-1}{2}$ 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 $\displaystyle c=\pm\frac{a}{2}$ is cleared.

We can safely assume now that \[|c| \le \frac{a-1}{2}.\] Automatically, $|d| \le \frac{b-1}{2}$, since \[2|d| = 2|\frac{1-bc}{a} | \le \frac{b(a-1)+2}{a} < \frac{ba}{a} = b,\] since $b>a>1$ implies $b>2$.