Difference between revisions of "1988 IMO Problems/Problem 6"
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We construct the following inequalities: <math>r_2=\frac{b^2-c}{a}<a,</math> since <math>c</math> is positive. Adding <math>b</math>, <math>r_2+b<a+b,</math> contradicting the minimality of <math>a+b.</math> | We construct the following inequalities: <math>r_2=\frac{b^2-c}{a}<a,</math> since <math>c</math> is positive. Adding <math>b</math>, <math>r_2+b<a+b,</math> contradicting the minimality of <math>a+b.</math> | ||
− | -Benedict T | + | -Benedict T (countmath1) |
Revision as of 00:06, 25 December 2022
Problem
Let and
be positive integers such that
divides
. Show that
is the square of an integer.
Solution 1
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.
Solution 2 (Sort of Root Jumping)
We proceed by way of contradiction.
WLOG, let fix
, and choose the value of
, such that
is minimized. Let
be the nonsquare positive integer such that such that
or
Expanding and rearranging,
This quadratic has two roots,
and
, such that
WLOG, let
. By Vieta's,
and
From
,
is an integer, because both
and
are integers.
From
is nonzero since
is not square, from our assumption.
We can plug in for
in the original expression, because
yielding
. If
then
and
and because
is a positive integer.
We construct the following inequalities: since
is positive. Adding
,
contradicting the minimality of
-Benedict T (countmath1)
1988 IMO (Problems) • Resources | ||
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