Difference between revisions of "1988 IMO Problems/Problem 6"
Countmath1 (talk | contribs) (→Solution 2 (Sort of Root Jumping)) |
Countmath1 (talk | contribs) (→Solution 2 (Sort of Root Jumping)) |
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We proceed by way of contradiction. | We proceed by way of contradiction. | ||
− | WLOG, let <math>a\geq{b},</math> fix <math>b</math>, and choose the value of <math>a</math>, such that <math>a+b</math> is minimized. | + | WLOG, let <math>a\geq{b},</math> fix <math>b</math>, and choose the value of <math>a</math>, such that <math>a+b</math> is minimized. Fix <math>c</math> to be the nonsquare positive integer such that such that <math>\frac{a^2+b^2}{ab+1}=c,</math> or <math>a^2+b^2=c(ab+1).</math> Expanding and rearranging, |
<cmath>P(a)=a^2+a(-bc)+b^2-c=0.</cmath> | <cmath>P(a)=a^2+a(-bc)+b^2-c=0.</cmath> | ||
This quadratic has two roots, <math>r_1</math> and <math>r_2</math>, such that | This quadratic has two roots, <math>r_1</math> and <math>r_2</math>, such that |
Revision as of 00:08, 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. Fix
to 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)
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