Difference between revisions of "1988 IMO Problems/Problem 6"

Revision as of 10:39, 10 August 2023

Problem

Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^{2} + b^{2}$ . Show that $\frac {a^{2} + b^{2}}{ab + 1}$ is the square of an integer.

Solution 1

Choose integers $a,b,k$ such that $a^2+b^2=k(ab+1)$ Now, for fixed $k$ , out of all pairs $(a,b)$ choose the one with the lowest value of $\min(a,b)$ . Label $b'=\min(a,b), a'=\max(a,b)$ . Thus, $a'^2-kb'a'+b'^2-k=0$ is a quadratic in $a'$ . Should there be another root, $c'$ , the root would satisfy: $b'c'\leq a'c'=b'^2-k<b'^2\implies c'<b'$ Thus, $c'$ isn't a positive integer (if it were, it would contradict the minimality condition). But $c'=kb'-a'$ , so $c'$ is an integer; hence, $c'\leq 0$ . In addition, $(a'+1)(c'+1)=a'c'+a'+c'+1=b'^2-k+b'k+1=b'^2+(b'-1)k+1\geq 1$ so that $c'>-1$ . We conclude that $c'=0$ so that $b'^2=k$ .

This construction works whenever there exists a solution $(a,b)$ for a fixed $k$ , hence $k$ is always a perfect square.

Solution 2 (Sort of Root Jumping)

We proceed by way of contradiction.

WLOG, let $a\geq{b}$ and fix $c$ to be the nonsquare positive integer such that such that $\frac{a^2+b^2}{ab+1}=c,$ or $a^2+b^2=c(ab+1).$ Choose a pair $(a, b)$ out of all valid pairs such that $a+b$ is minimized. Expanding and rearranging, $P(a)=a^2+a(-bc)+b^2-c=0.$ This quadratic has two roots, $r_1$ and $r_2$ , such that $(a-r_1)(a-r_2)=P(a)=0.$ WLOG, let $r_1=a$ . By Vieta's, $\textbf{(1) } r_2=bc-a,$ and $\textbf{(2) } r_2=\frac{b^2-c}{a}.$ From $\textbf{(1)}$ , $r_2$ is an integer, because both $b$ and $c$ are integers.

From $\textbf{(2)},$ $r_2$ is nonzero since $c$ is not square, from our assumption.

We can plug in $r_2$ for $a$ in the original expression, because $P(r_2)=P(a)=0,$ yielding $c=\frac{r^2_2+b^2}{r_2b+1}$ . If $c>0,$ then $r_2b+1>0,$ and $r_2b+1\neq{0},$ and because $b>0, r_2$ is a positive integer.

We construct the following inequalities: $r_2=\frac{b^2-c}{a}<a,$ since $c$ is positive. Adding $b$ , $r_2+b<a+b,$ contradicting the minimality of $a+b.$

-Benedict T (countmath1)

Video Solutions

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

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	-Benedict T (countmath1)		-Benedict T (countmath1)
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	==Video Solutions==		==Video Solutions==

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