Eötvös 1917/1 solution

24th Eötvös 1917

Problem 1

a, b are integers. The solutions of y - 2x = a, y² - xy + x² = b are rational. Show that they must be integers.

Solution

We have x = (y-a)/2, substituting in the other equation gives 3y² = 4b-a². Hence (3y)² = 3(4b-a²), which is an integer, so 3y is an integer. Hence 3(4b-a²) is a square, so (4b-a²) must be divisible by 3. Hence y² = (4b-a²)/3 is an integer, so y is an integer.

Since 3y² + a² = 4b, which is even, y and a must have the same parity. Hence y-a is an even integer, and so x is an integer.

Note that substituting y = a+2x does not work so easily.

24th Eötvös 1917