# Difference between revisions of "2008 IMO Problems/Problem 3"

## 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

Solution 1

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 $n$ such that $p$ divides $n^2+1$ and $p>2n+\sqrt{2n}$. 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 hypothesis.

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 is a good choice. For starters, notice that $(a^2+b^2)(c^2+d^2)=n^2+1$.

If $c=\pm\frac{a}{2}$, from (1) we see that $a$ must divide $2$ and hence $a=2$. This implies, $d=-\frac{b-1}{2}$ and $n=\frac{b(b-1)}{2}-2$. Therefore, $\left(b-\frac{1}{2}\right)^2 = 1/4 + 2(n+2) > 2n$, so $b > \sqrt{2n}+\frac{1}{2}$. 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 $$|d| \le \frac{b-1}{2}.$$

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

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 is not hard to show for large enough $p$.

Now armed with $a+b-1>\sqrt{p}$ and (2), we get

$4(n^2+1) & \le p( a^2+b^2 - 2(a+b-1) ) \\ & < p( p-2\sqrt{p} ) \\ & < u^2(u-1)^2,$ (Error compiling LaTeX. ! Misplaced alignment tab character &.)

where $u=\sqrt{p}.$

Finally, $$u^2(u-1)^2 > 4n^2+4 > 4n^2\Rightarrow \\ u(u-1) > 2n \Rightarrow \\ (u-\frac{1}{2})^2 > 2n+\frac{1}{4} > 2n \Rightarrow \\ u > \sqrt{2n} + \frac{1}{2} \Rightarrow \\ p = u^2 > 2n + \sqrt{2n}.$$ The proof is complete.--Vbarzov 03:02, 5 September 2008 (UTC)

Solution 2

We begin with a lemma.

Lemma: For every prime $p \equiv 1 \mod{4}$, there exists a positive integer $n \le \dfrac{p - \sqrt{p-4}}{2}$ such that $p | n^2 + 1$.

Proof: We know that there must be a solution $x$ to the equation $$x^2+1 \equiv 0 \mod{p}$$ with $1 \le x \le p-1$. However, if $x$ is a solution then so is $p-x$, so there must be a solution $x \le p-1$. Let $n$ denote this solution and suppose that $n = \frac{p-k}{2}$. Then, we have $$\left(\dfrac{p-k}{2}\right)^2 \equiv -1 \mod{4}$$ $$k^2 \equiv -4 \mod{4}.$$ Since $k$ is a positive integer, it follows that $k \ge p-4$, so we have $$n \le \dfrac{p - \sqrt{p-4}}{2},$$ as desired. The lemma is proven.

Suppose for sake of contradiction that there are only finitely many integers $n_1,n_2,\dots,n_k$ that work. Let $p \equiv 1 \mod{4}$ be a prime with $p>20$ and such that $p$ is relatively prime to $n_i^2+1$ for all $i$ (the existence of $p$ is guaranteed by the existence of infinitely many primes of the form $4k+1$). Then, we know that there exists an $N$ such that $N\le \dfrac{p - \sqrt{p-4}}{2}$ and $p | N^2 + 1$ (this last condition shows that $N$ is not among $n_1,n_2,\dots,n_k$. We want to show that $$p > 2N + \sqrt{2N}$$ $$p - 2N > \sqrt{2N}.$$ But, we know that $p-2N \ge \sqrt{p-4}$, so it suffices to show that $$p-4 > 2N$$ $$p - 2N > 4.$$ Once again, it suffices to show that $$\sqrt{p-4} > 4,$$ which follows from $p>20$.

Thus, $N$ satisfies the required condition and it follows that there exist infinitely many values of $n$ that satisfy the given condition, as desired.