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

(→‎Solution 2 (Sort of Root Jumping))
(→‎Solution 2 (Sort of Root Jumping))
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We proceed by way of contradiction.
 
We proceed by way of contradiction.
  
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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,  
+
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 23:08, 24 December 2022

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 $a\geq{b},$ fix $b$, and choose the value of $a$, such that $a+b$ is minimized. Fix $c$ to be the nonsquare positive integer such that such that $\frac{a^2+b^2}{ab+1}=c,$ or $a^2+b^2=c(ab+1).$ 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)



1988 IMO (Problems) • Resources
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