## 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*=*b*^{2}+*b*+1*. Prove that the equation*

*ax*^{2}-(2*b*+1)*xy*+*cy*^{2}=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}>*b*^{2}+*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*-2*b*+*c*-1. These are integers, satisfying

B^{2}+B+1 =(b-a)^{2}+(b-a)+1=

= (b^{2}+b+1)-2ab+a^{2}-a=

= ac-2ab+a^{2}-a=a(a-2b+c-1)=AC,

and furthermore, *A*>0, 0\(\displaystyle \le\)*B*<*b*, and *C*=(*B*^{2}+*B*+1)/*A*>0. Hence the induction hypothesis applied to the numbers *A*, *B*, *C*, yields integers *X* and *Y*, such that

*AX*^{2}-(2*B*+1)*XY*+*CY*^{2}=1.

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

1 = AX^{2}-(2B+1)XY+CY^{2}=

= aX^{2}-(2b-2a+1)XY+(a-2b+c-1)Y^{2}=

= a(X+Y)^{2}-(2b+1)(X+Y)Y+cY^{2}.

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 *ax*^{2}-(2*b*+1)*xy*+*cy*^{2}<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 *ax*^{2}-(2*b*+1)*xy*+*cy*^{2} 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 *x*^{2}- *xy*+*y*^{2}<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.