### 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 \( \displaystyle \frac{\delta}{\gamma} \).