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Five in a Row

Euler did not use matrices as we might today to find the coefficients in the general second degree equation \( \alpha yy + \beta xy + \gamma xx + \delta y + \epsilon x + \zeta = 0 \). He simply substituted values and manipulated equations. Substituting \( x=y=0 \) into the equation gives \( \zeta=0 \). The points \( (1,1) \) and \( (2,2) \) give rise to the equations \begin{eqnarray*} \alpha + \beta + \gamma + \delta + \epsilon &=& 0\\ 4\alpha + 4\beta + 4\gamma + 2\delta + 2\epsilon &=& 0.\\ \end{eqnarray*} These can be simplified to \begin{eqnarray*} \alpha + \beta + \gamma &=& 0\\ \delta + \epsilon &=& 0,\\ \end{eqnarray*} which in turn give \( \beta = -(\alpha + \gamma) \) and \( \epsilon = -\delta \). Substituting these back into the general second degree equation gives us \begin{equation} \alpha y^2 - (\alpha + \gamma)xy + \gamma x^2 + \delta y - \delta x = 0, \end{equation} as Euler observed in his letter. The problem is that when \( (3,3) \) or \( (4,4) \) (or any point of the form \( (k,k) \) ) is substituted in this equation, the result is \( 0=0 \). Thus, in modern terms, the five points give rise to a system only of rank 3, and there is no unique solution.

The previous equation can be factored as follows \begin{equation} (y-x)(\alpha y - \gamma x + \delta) = 0. \label{FactoredConic} \end{equation} Because a product can only be zero if one of its factors is zero, this in turn means that \[ y-x=0 \quad \mbox{or} \quad \alpha y - \gamma x + \delta=0. \] The first of these equations is the equation of the line \( y=x \), which contains all of the given points. The other equation is a completely arbitrary linear equation, which could even be \( y=x \). As long as \( \alpha \ne 0 \), it can be re-written as \[ y = mx + b, \quad \mbox{where} \quad m = \frac{\gamma}{\alpha} \quad \mbox{and} \quad b = -\frac{\delta}{\alpha}. \] This slope and intercept can be thought of as the "two coefficients to be determined" that Euler mentioned and the factored equation above becomes \begin{equation} (y-x)(y-mx-b)=0. \label{SlopeInterceptConic} \end{equation} We note that if \( \alpha = 0 \) in the factored equation, then the second equation is of a vertical line. In this case, the two coefficients in question are \( \alpha = 0 \), and the \( x \)-value \( \displaystyle \frac{\delta}{\gamma} \).