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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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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Revision as of 20:56, 25 October 2007
Problem
Let and
be positive integers such that
divides
. Show that
is the square of an integer.
Solution
Choose integers such that
Now, for fixed
, out of all pairs
choose the one with the lowest value of
. Label
. Thus,
is a quadratic in
. Should there be another root,
, the root would satisfy:
Thus,
isn't a positive integer (if it were, it would contradict the minimality condition). But
, so
is an integer; hence,
. In addition,
so that
. We conclude that
so that
.
This construction works whenever there exists a solution for a fixed
, hence
is always a perfect square.
1988 IMO (Problems) • Resources | ||
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