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Difference between revisions of "2008 IMO Problems/Problem 3"
Line 33: Line 33:
 
Now armed with <math>a+b-1>\sqrt{p}</math> and (2), we get
 
Now armed with <math>a+b-1>\sqrt{p}</math> and (2), we get
 
<cmath>
 
<cmath>
−
\begin{align}
+
4(n^2+1) \le p( a^2+b^2 - 2(a+b-1) ) < \\
−
4(n^2+1) & \le p( a^2+b^2 - 2(a+b-1) ) < \\
+
          < p( p-2\sqrt{p} ) < \\
−
        & < p( p-2\sqrt{p} ) < \\
+
          < u^2(u-1)^2,
−
        & < u^2(u-1)^2,
 
−
\end{align}
 
 
</cmath>
 
</cmath>
 
where <math>u=\sqrt{p}.</math>
 
where <math>u=\sqrt{p}.</math>

Revision as of 23:33, 3 September 2008

Problem

Prove that there are infinitely many positive integers $n$ such that $n^{2} + 1$ has a prime divisor greater than $2n + \sqrt {2n}$.

Solution

The main idea is to take a gaussian prime $a+bi$ and multiply it by a "twice as small" $c+di$ 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 what we are looking for, and $p=n^2+1 > 2n+\sqrt{2n}$ as long as $p$ (and hence $n$) is large enough. Assume from now on that $b>a>1$.

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

Define $n=|ac-bd|$ and let's see why this was a good choice. For starters, notice that $(a^2+b^2)(c^2+d^2)=n^2+1$.


If $c=\pm\frac{a}{2}$, then from (1), we see that $a$ must divide $2$ and hence $a=2$. In turn, $d=-\frac{b-1}{2}$ 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 \[|c| \le \frac{a-1}{2}.\] As $b>a>1$ implies $b>2$, we have \[|d| = \left|\frac{1-bc}{a}\right| \le \frac{b(a-1)+2}{2a} < \frac{ba}{2a} = \frac{b}{2},\] so \[|d| \le \frac{b-1}{2}.\]

\[n^2+1 = (a^2+b^2)(c^2+d^2) \le p\left( \frac{(a-1)^2}{4}+\frac{(b-1)^2}{4} \right). \quad (2)\]

Before we proceed, we would like to show first that $a+b-1 > \sqrt{p}$. Observe that the function $x+\sqrt{p-x^2}$ over $x\in(2,\sqrt{p-4})$ reaches its minima on the ends, so $a+b$ given $a^2+b^2=p$ is minimized for $a = 2$, where it equals $2+\sqrt{p-2^2}$. So we want to show that \[2+\sqrt{p-4} > \sqrt{p} + 1,\] which obviously holds for large $p$.

Now armed with $a+b-1>\sqrt{p}$ and (2), we get \[4(n^2+1) \le p( a^2+b^2 - 2(a+b-1) ) < \\ < p( p-2\sqrt{p} ) < \\ < u^2(u-1)^2,\] where $u=\sqrt{p}.$

Finally, \[u^2(u-1)^2 > 4n^2+4 > 4n^2\Rightarrow \\ u(u-1) > 2n \Rightarrow u > \sqrt{2n} + \frac{1}{2} \Rightarrow \\ p = u^2 > 2n + \sqrt{2n}.\]

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