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Difference between revisions of "1988 IMO Problems/Problem 6"

(New page: Let <math>a</math> and <math>b</math> be positive integers such that <math>ab + 1</math> divides <math>a^{2} + b^{2}</math>. Show that <math> \frac {a^{2} + b^{2}}{ab + 1} </math> is the s...)
 
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==Problem==
 
Let <math>a</math> and <math>b</math> be positive integers such that <math>ab + 1</math> divides <math>a^{2} + b^{2}</math>. Show that
 
Let <math>a</math> and <math>b</math> be positive integers such that <math>ab + 1</math> divides <math>a^{2} + b^{2}</math>. Show that
 
<math>
 
<math>
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This construction works whenever there exists a solution <math>(a,b)</math> for a fixed <math>k</math>, hence <math>k</math> is always a perfect square.
 
This construction works whenever there exists a solution <math>(a,b)</math> for a fixed <math>k</math>, hence <math>k</math> is always a perfect square.
  
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==References==
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{{IMO box|year=1988|num-b=5|after=Last question}}
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*http://www.kalva.demon.co.uk/imo/isoln/isoln886.html
 

Revision as of 19:56, 25 October 2007

Problem

Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^{2} + b^{2}$. Show that $\frac {a^{2} + b^{2}}{ab + 1}$ is the square of an integer.

Solution

Choose integers $a,b,k$ such that $a^2+b^2=k(ab+1)$ Now, for fixed $k$, out of all pairs $(a,b)$ choose the one with the lowest value of $\min(a,b)$. Label $b'=\min(a,b), a'=\max(a,b)$. Thus, $a'^2-kb'a'+b'^2-k=0$ is a quadratic in $a'$. Should there be another root, $c'$, the root would satisfy: $b'c'\leq a'c'=b'^2-k<b'^2\implies c'<b'$ Thus, $c'$ isn't a positive integer (if it were, it would contradict the minimality condition). But $c'=kb'-a'$, so $c'$ is an integer; hence, $c'\leq 0$. In addition, $(a'+1)(c'+1)=a'c'+a'+c'+1=b'^2-k+b'k+1=b'^2+(b'-1)k+1\geq 1$ so that $c'>-1$. We conclude that $c'=0$ so that $b'^2=k$.

This construction works whenever there exists a solution $(a,b)$ for a fixed $k$, hence $k$ is always a perfect square.

1988 IMO (Problems) • Resources
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