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== | + | ==Video Solution== |
+ | https://youtu.be/wqCdEE1Ueh0 | ||
+ | |||
+ | ==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 16: | ||
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> 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> | ||
+ | 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 (countmath1) | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=usEQRx4J_ew ~KevinChen_Yay | ||
+ | |||
+ | ==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. | ||
{{IMO box|year=1988|num-b=5|after=Last question}} | {{IMO box|year=1988|num-b=5|after=Last question}} |
Latest revision as of 19:24, 25 September 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
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.
1988 IMO (Problems) • Resources | ||
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