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Difference between revisions of "2008 IMO Problems/Problem 3"
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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.
  
−
Take a prime <math>p</math> of the form <math>4k+1</math> and consider its "sum-of-two squares" representation <math>p=a^2+b^2</math>, which we know to exist for all such primes. If <math>a=1</math> or <math>b=1</math>, then <math>n=b</math> or <math>n=a</math> is our guy, and <math>p=n^2+1 > 2n+\sqrt{2n}</math> as long as <math>p</math> (and hence <math>n</math>) is large enough. Let's see what happens when both <math>a>1</math> and <math>b>1</math>. Apparently, <math>a\ne b</math>, so assume without loss of generality that <math>b>a>1</math>.
+
Take a prime <math>p</math> of the form <math>4k+1</math> and consider its "sum-of-two squares" representation <math>p=a^2+b^2</math>, which we know to exist for all such primes. As <math>a\ne b</math>, assume without loss of generality that <math>b>a</math>. If <math>a=1</math>, then <math>n=b</math> is our guy, and <math>p=n^2+1 > 2n+\sqrt{2n}</math> as long as <math>p</math> (and hence <math>n</math>) is large enough. Let's see what happens when <math>b>a>1</math>.
  
 
Since <math>a</math> and <math>b</math> are (obviously) co-prime, there must exist integers <math>c</math> and <math>d</math> such that
 
Since <math>a</math> and <math>b</math> are (obviously) co-prime, there must exist integers <math>c</math> and <math>d</math> such that
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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>\displaystyle c=\pm\frac{a}{2}</math> is cleared.
+
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>\displaystyle 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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Automatically, <math>|d| \le \frac{b-1}{2}</math>, since  
 
Automatically, <math>|d| \le \frac{b-1}{2}</math>, since  
 
<cmath>2|d| = 2|\frac{1-bc}{a} | \le \frac{b(a-1)+2}{a} < \frac{ba}{a} = b,</cmath>
 
<cmath>2|d| = 2|\frac{1-bc}{a} | \le \frac{b(a-1)+2}{a} < \frac{ba}{a} = b,</cmath>
−
since <math>b>a>1</math> implies <math>b>2</math>.
+
as <math>b>a>1</math> implies <math>b>2</math>.
 +
We have
 +
<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>

Revision as of 21:37, 3 September 2008

(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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