Difference between revisions of "1988 IMO Problems/Problem 6"
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</math> | </math> | ||
is the square of an integer. | is the square of an integer. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/wqCdEE1Ueh0 | ||
==Solution 1== | ==Solution 1== | ||
Line 17: | Line 20: | ||
We proceed by way of contradiction. | We proceed by way of contradiction. | ||
− | WLOG, let <math>a\geq{b} | + | WLOG, let <math>a\geq{b}</math> and 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> Choose a pair <math>(a, b)</math> out of all valid pairs such that <math>a+b</math> is minimized. 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 | ||
Line 34: | Line 37: | ||
-Benedict T (countmath1) | -Benedict T (countmath1) | ||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=usEQRx4J_ew ~KevinChen_Yay | ||
+ | |||
+ | |||
+ | {{IMO box|year=1988|num-b=5|after=Last question}} | ||
+ | ==Solution 3== | ||
+ | Given that <math>ab+1</math> divides <math>a^2+b^2</math>, we have <math>a^2+b^2=k(ab+1)</math> for some integer <math>k</math>. | ||
+ | |||
+ | Expanding the right side, we get <math>a^2+b^2=kab+k</math>. Rearranging terms, we have <math>a^2-kab+b^2-k=0</math>. | ||
+ | |||
+ | Consider this as a quadratic equation in <math>a</math>. By the quadratic formula, we have | ||
+ | <cmath>a=\frac{kb\pm\sqrt{k^2b^2-4(b^2-k)}}{2}.</cmath> | ||
+ | |||
+ | For <math>a</math> to be an integer, the discriminant <math>k^2b^2-4(b^2-k)</math> must be a perfect square. Let <math>k^2b^2-4(b^2-k)=m^2</math> for some integer <math>m</math>. | ||
+ | Rearranging terms, we get <math>m^2=k^2b^2-4b^2+4k</math>. Factoring the right side, we have <math>m^2=(kb-2)^2</math>. | ||
+ | Thus, <math>m=kb-2</math> and <math>a=\frac{kb\pm(kb-2)}{2}=b</math> or <math>a=kb-b</math>. In either case, we have <math>a=kb-b</math>. | ||
+ | Substitute <math>a=kb-b</math> back into <math>a^2+b^2=kab+k</math>, we get <math>b^2+k^2b^2-2kb^2+b^2=kb^2-kb+k</math>. | ||
− | {{ | + | Simplifying, we have <math>b^2=k</math>. Therefore, <math>\frac{a^2+b^2}{ab+1}=\frac{(kb-b)^2+b^2}{b(k)+1}=\frac{b^2}{b+1}=b</math>, which is the square of an integer. |
+ | By M. Nazaryan. |
Latest revision as of 19:50, 24 August 2024
Contents
Problem
Let and
be positive integers such that
divides
. Show that
is the square of an integer.
Video Solution
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 and fix
to be the nonsquare positive integer such that such that
or
Choose a pair
out of all valid pairs such that
is minimized. 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)
Video Solution
https://www.youtube.com/watch?v=usEQRx4J_ew ~KevinChen_Yay
1988 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last question |
All IMO Problems and Solutions |
Solution 3
Given that divides
, we have
for some integer
.
Expanding the right side, we get . Rearranging terms, we have
.
Consider this as a quadratic equation in . By the quadratic formula, we have
For to be an integer, the discriminant
must be a perfect square. Let
for some integer
.
Rearranging terms, we get . Factoring the right side, we have
.
Thus, and
or
. In either case, we have
.
Substitute back into
, we get
.
Simplifying, we have . Therefore,
, which is the square of an integer.
By M. Nazaryan.