Crowdly

Додати до Chrome

Problem 6 in the 29th International Mathematical Olympiad (1988) is considered o...

✅ Перевірена відповідь на це питання доступна нижче. Наші рішення, перевірені спільнотою, допомагають краще зрозуміти матеріал.

Problem 6 in the 29th International Mathematical Olympiad (1988) is considered one of the hardest problems in IMO.

IMO-29 Probem 6: Let a and b be positive integers such that ab+1 divides a^2+b^2. Show that $\frac{a^2+b^2}{ab+1}$ \frac{a^2+b^2}{ab+1} is the square of an integers.

To prove this problem, we first describe the problem using the techniques we have learned in this course.

1) Define a function $f: N^2 \to Q$ f: N^2 \to Q with $f(x,y) = \frac{x^2+y^2}{xy+1}$ f(x,y) = \frac{x^2+y^2}{xy+1}.

2) Let Z^+ be the domain of variables a, b, and k, prove the following logic statement:

$\forall a \forall b [(f(a,b) \in Z^+) \to (\exists k (f(a, b)=k^2))]$ \forall a \forall b [(f(a,b) \in Z^+) \to (\exists k (f(a, b)=k^2))]

Next, choose the correct step for each statement in the formal proof.

Case (1): a=b.

In this case, we let c = lb-a and we will first derive the upper bound of c.

Since l=f(x,0)=x^2, k=x exists.

Case (2): $a\not=b$ a\not=b. Without loss of generality, we consider a>b.

Next, we derive the lower bound of c.

Simplifying the inequality above, we have a+b>lb and the upper bound of c is c < b.

Simplifying the equation above, we have l(bc+1)=b^2+c^2>0 so we find that f(b,c)=l and the lower bound of c is $0 \leq c$ 0 \leq c.

Using the direct proof technique, we assume that $l=f(a,b) \in Z^+$ l=f(a,b) \in Z^+.

When l=f(a,b) for a pair of positive integers a and b, there are two cases.

In this case, $f(a,b)=f(a,a)=\frac{2a^2}{a^2+1}=l$ f(a,b)=f(a,a)=\frac{2a^2}{a^2+1}=l.

Repeating the steps from Step 11, we can find a sequence of positive integers $a>b>c>\cdots > x>0$ a>b>c>\cdots > x>0 such that $f(a,b)=f(b,c)=\cdots=f(x,0)=l$ f(a,b)=f(b,c)=\cdots=f(x,0)=l.