Difference between revisions of "1988 IMO Problems/Problem 6"
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is the square of an integer. | is the square of an integer. | ||
− | ==Solution== | + | ==Solution 1== |
Choose integers <math>a,b,k</math> such that <math>a^2+b^2=k(ab+1)</math> | Choose integers <math>a,b,k</math> such that <math>a^2+b^2=k(ab+1)</math> | ||
Line 13: | Line 13: | ||
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. | ||
+ | |||
+ | ==Solution 2 (Sort of Root Jumping)== | ||
+ | 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. Let <math>c</math> 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> | ||
+ | This quadratic has two roots, <math>r_1</math> and <math>r_2</math>, such that | ||
+ | <cmath>(a-r_1)(a-r_2)=P(a)=0.</cmath> | ||
+ | WLOG, let <math>r_1=a</math>. By Vieta's, | ||
+ | <math>\textbf{(1) } r_2=bc-a,</math> and | ||
+ | <math>\textbf{(2) } r_2=\frac{b^2-c}{a}.</math> | ||
+ | From <math>\textbf{(1)}</math>, <math>r_2</math> is an integer, because both <math>b</math> and <math>c</math> are integers. | ||
+ | |||
+ | From <math>\textbf{(2)},</math> <math>r_2</math> is nonzero since <math>c</math> is not square, from our assumption. | ||
+ | |||
+ | We can plug in <math>r_2</math> for <math>a</math> in the original expression, because <math>P(r_2)=P(a)=0,</math> yielding <math>c=\frac{r^2_2+b^2}{r_2b+1}</math>. If <math>c>0,</math> then <math>r_2b+1>0,</math> and <math>r_2b+1\neq{0},</math> and because <math>b>0, r_2</math> is a positive integer. | ||
+ | |||
+ | 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 | ||
+ | |||
+ | |||
+ | |||
+ | |||
{{IMO box|year=1988|num-b=5|after=Last question}} | {{IMO box|year=1988|num-b=5|after=Last question}} |
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
1988 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last question |
All IMO Problems and Solutions |