### 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 $$\alpha y^2 - (\alpha + \gamma)xy + \gamma x^2 + \delta y - \delta x = 0,$$ 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 $$(y-x)(\alpha y - \gamma x + \delta) = 0. \label{FactoredConic}$$ 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 $$(y-x)(y-mx-b)=0. \label{SlopeInterceptConic}$$ 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 $$\displaystyle \frac{\delta}{\gamma}$$.