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English Issue, December 2002

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Solutions of advanced problems A

A. 274. Let a, b, c be positive integers for which ac=b2+b+1. Prove that the equation

ax2-(2b+1)xy+cy2=1

has an integer solution.

Polish competition problem

Solution 1. The proof is by induction on b. If b=0, then a=c=1, and x=y=1 is a solution.

Now suppose that b is positive, and the statement is true for all smaller values of b. The numbers a and c cannot be both greater than b, since if a,c\(\displaystyle \ge\)b+1, then ac\(\displaystyle \ge\)(b+1)2>b2+b+1. As the statement remains the same interchanging a and c, we can assume, without loss of generality, that a\(\displaystyle \le\)b.

Let A=a, B=b-a, C=a-2b+c-1. These are integers, satisfying

B2+B+1 =(b-a)2+(b-a)+1=
= (b2+b+1)-2ab+a2-a=
= ac-2ab+a2-a=a(a-2b+c-1)=AC,

and furthermore, A>0, 0\(\displaystyle \le\)B<b, and C=(B2+B+1)/A>0. Hence the induction hypothesis applied to the numbers A, B, C, yields integers X and Y, such that

AX2-(2B+1)XY+CY2=1.

By substituting the definitions of A, B and C, we get

1 = AX2-(2B+1)XY+CY2=
= aX2-(2b-2a+1)XY+(a-2b+c-1)Y2=
= a(X+Y)2-(2b+1)(X+Y)Y+cY2.

Therefore, the number pair x=X-Y, y=Y is a solution of the equation.

Solution 2. Consider the set of points in the Cartesian system for which ax2-(2b+1)xy+cy2<2. Simple calculation shows that they form an ellipse whose area is \(\displaystyle \frac{4\pi}{\sqrt3}\) units.

The elliptical disc is symmetric about the origin, it is convex, and its area is greater than 4 units. Thus, according to Minkowski's theorem there is a lattice point other than the origin in its interior. At that point, the value of ax2-(2b+1)xy+cy2 is between 0 and 2, therefore it is 1.

Remark. In Solution 1, we are actually using the area preserving transformation (x,y)\(\displaystyle \mapsto\)(x-y,y) to map the elliptical disc onto another ellipse of equal area with smaller coefficients. By repeating the transformation several times, we will always obtain the ellipse x2- xy+y2<2. This ellipse contains six lattice points different from the origin, they are (\(\displaystyle \pm\)1,0), (0,\(\displaystyle \pm\)1) and (\(\displaystyle \pm\)1,\(\displaystyle \pm\)1). Therefore, the equation always has six solutions.

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