## Solution: Day 1, problem 3

Let

*m*be the quotient. We may suppose that (

*a*,

*b*) is a minimal positive solution of the equation (

*a*

^{2}+

*b*

^{2}) =

*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*

*a*

^{2}=

*bb*' +

*m*. If

*b*' is positive, then this equation implies

*b*' = (

*a*

^{2}-

*m*)/

*b*<

*a*

^{2}/

*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*=

*a*

^{2}-

*bb*' ≥

*a*

^{2}+

*b*>

*b*>

*ma*≥

*m*, a contradiction. Hence

*b*' = 0 and

*m*=

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