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}(2b+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=ba, C=a2b+c1. These are integers, satisfying
B^{2}+B+1 =(ba)^{2}+(ba)+1= = (b^{2}+b+1)2ab+a^{2}a= = ac2ab+a^{2}a=a(a2b+c1)=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}(2B+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}(2b2a+1)XY+(a2b+c1)Y^{2}= = a(X+Y)^{2}(2b+1)(X+Y)Y+cY^{2}.
Therefore, the number pair x=XY, y=Y is a solution of the equation.
Solution 2. Consider the set of points in the Cartesian system for which ax^{2}(2b+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}(2b+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)\(\displaystyle \mapsto\)(xy,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.
