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Difference between revisions of "2008 IMO Problems/Problem 3"
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(still editing...)
 
(still editing...)
  
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The main idea is to take a gaussian prime <math>a+bi</math> and multiply it by a "twice smaller" <math>c+di</math> to get <math>n+i</math>. The rest is just making up the little details.
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The main idea is to take a gaussian prime <math>a+bi</math> and multiply it by a "twice as small" <math>c+di</math> to get <math>n+i</math>. The rest is just making up the little details.
  
 
For each sufficiently large prime <math>p</math> of the form <math>4k+1</math>, we shall find a corresponding <math>n</math> satisfying the required condition with the prime number in question being <math>p</math>. Since there exist infinitely many such primes and, for each of them, <math>n \ge \sqrt{p-1}</math>, we will have found infinitely many distinct <math>n</math> satisfying the problem.
 
For each sufficiently large prime <math>p</math> of the form <math>4k+1</math>, we shall find a corresponding <math>n</math> satisfying the required condition with the prime number in question being <math>p</math>. Since there exist infinitely many such primes and, for each of them, <math>n \ge \sqrt{p-1}</math>, we will have found infinitely many distinct <math>n</math> satisfying the problem.
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If <math>c=\pm\frac{a}{2}</math>, then from (1), we get <math>a\2</math> and hence <math>a=2</math>. That means that <math>d=-\frac{b-1}{2}</math> and <math>n=\frac{b(b-1)}{2}-2</math>. Therefore, <math>b^2-b=2n+4>2n</math> and so <math>\left(b-\frac{1}{2}\right)^2>2n</math>, from where <math>b > \sqrt{2n}+\frac{1}{2}</math>. Finally, <math>p=b^2+2^2 > 2n+\sqrt{2n}</math> and the case <math>c=\pm\frac{a}{2}</math> is cleared.
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If <math>c=\pm\frac{a}{2}</math>, then from (1), we get <math>a\\2</math> and hence <math>a=2</math>. That means that <math>d=-\frac{b-1}{2}</math> and <math>n=\frac{b(b-1)}{2}-2</math>. Therefore, <math>b^2-b=2n+4>2n</math> and so <math>\left(b-\frac{1}{2}\right)^2>2n</math>, from where <math>b > \sqrt{2n}+\frac{1}{2}</math>. Finally, <math>p=b^2+2^2 > 2n+\sqrt{2n}</math> and the case <math>c=\pm\frac{a}{2}</math> is cleared.
  
 
We can safely assume now that
 
We can safely assume now that
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<cmath>|d| \le \frac{b-1}{2}.</cmath>
 
<cmath>|d| \le \frac{b-1}{2}.</cmath>
 
<cmath>n^2+1 = (a^2+b^2)(c^2+d^2) > p \left( \frac{(a-1)^2}{4}+\frac{(b-1)^2}{4} \right).</cmath>
 
<cmath>n^2+1 = (a^2+b^2)(c^2+d^2) > p \left( \frac{(a-1)^2}{4}+\frac{(b-1)^2}{4} \right).</cmath>
 +
Before we proceed, we would like to show that <math>a+b-1 > \sqrt{p}</math>. Observe that the function <math>x+\sqrt{p-x^2}</math> reaches its minima on <math>x\in(0,\sqrt{p})</math> on the edges, so <math>a+b</math> given <math>a^2+b^2=p</math> is minimized for <math>a = 2</math>.

Revision as of 22:32, 3 September 2008

(still editing...)

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 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 $c\pm ma$ 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 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 $a\\2$ 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 $c=\pm\frac{a}{2}$ is cleared.

We can safely assume now that \[|c| \le \frac{a-1}{2}.\] As $b>a>1$, we have $b>2$ and so \[|d| = \left|\frac{1-bc}{a}\right| \le \frac{b(a-1)+2}{2a} < \frac{ba}{2a} = \frac{b}{2},\] so \[|d| \le \frac{b-1}{2}.\] \[n^2+1 = (a^2+b^2)(c^2+d^2) > p \left( \frac{(a-1)^2}{4}+\frac{(b-1)^2}{4} \right).\] Before we proceed, we would like to show that $a+b-1 > \sqrt{p}$. Observe that the function $x+\sqrt{p-x^2}$ reaches its minima on $x\in(0,\sqrt{p})$ on the edges, so $a+b$ given $a^2+b^2=p$ is minimized for $a = 2$.

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