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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>.
+
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>.
  
 
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
Line 13: Line 13:
 
Define <math>n=|ac-bd|</math> and let's see what happens. Notice that <math>(a^2+b^2)(c^2+d^2)=n^2+1</math>.
 
Define <math>n=|ac-bd|</math> and let's see what happens. Notice that <math>(a^2+b^2)(c^2+d^2)=n^2+1</math>.
  
−
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>\ds 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>\ds b > \sqrt{2n}+\frac{1}{2}</math>. Finally, <math>p=b^2+2^2 > 2n+\sqrt{2n}</math> and the case <math>\ds 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>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.
  
 
We can safely assume now that
 
We can safely assume now that
 
<cmath>|c| \le \frac{a-1}{2}.</cmath>
 
<cmath>|c| \le \frac{a-1}{2}.</cmath>
−
Automatically, <math>d \le \frac{b-1}{2}</math>.
+
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>
 +
since <math>b>a>1</math> implies <math>b>2</math>.

Revision as of 21:29, 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. 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 $b>2$.

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