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Solution: Day 1, problem 3

Let mm be the quotient. We may suppose that (a,b)(a, b) is a minimal positive solution of the equation (a2+b2)=m(ab+1)(a^{2} + b^{2} ) = m(ab + 1)
(i.e. one with the smallest value ofa+ba + b) for this value of mm. Without loss of generality, suppose that b≥a≥0b ≥ a ≥ 0, and set b′=ma−bb' = ma - b. Thena2=bb′+mThen a^{2} = bb' + m. If b′b' is positive, then this equation implies b′=(a2−m)/b<a2/b<bb' = (a^{2} - m)/b < a^{2}/b < b, and (a,b′)(a, b') is a smaller positive solution of the equation of which (a,b)(a, b) was supposed to be the minimal solution. If b′b' is negative, then m=a2−bb′≥a2+b>b>ma≥mm = a^{2} - bb' ≥ a^{2} + b > b > ma ≥ m, a contradiction. Hence b′=0b' = 0 and m=a2m = a^{2} .

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+y2x^{2} + mxy + y^{2}, which is the unique reduced quadratic form in its equivalence class.