Solution: Day 1, problem 3
Let m be the quotient. We may suppose that (a, b) is a minimal positive solution of the equation (a2 + b2 ) = m(ab + 1)
(i.e. one with the smallest value of a + b) for this value of m. Without loss of generality, suppose that b ≥ a ≥ 0, and set b' = ma - b. Then a2 = bb' + m. If b' is positive, then this equation implies b' = (a2 - m)/b < a2/b < b, and (a, b') is a smaller positive solution of the equation of which (a, b) was supposed to be the minimal solution. If b' is negative, then m = a2 - bb' ≥ a2 + b > b > ma ≥ m, a contradiction. Hence b' = 0 and m = a2 .
Note: This solution is just the explicit result of applying reduction theory (specifically, Sätze 1 and 2 of Section 13 of my book on quadratic fields) to the quadratic form x2 + mxy + y2 , which is the unique reduced quadratic form in its equivalence class.