Euler did not use matrices as we might today to find the coefficients in the general second degree equation \[\alpha y^2 + \beta xy + \gamma x^2 + \delta y + \varepsilon x + \zeta = 0.\] 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

\(\alpha + \beta + \gamma + \delta + \varepsilon\) |
\(=0\) |

\(4\alpha + 4\beta + 4\gamma + 2\delta + 2\varepsilon\) |
\(=0.\) |

These can be simplified to

\(\alpha + \beta + \gamma\) |
\(=0\) |

\(\delta + \varepsilon\) |
\(=0,\) |

which in turn give \(\beta = -(\alpha + \gamma)\) and \(\varepsilon = -\delta.\) Substituting these back into the general equation \[\alpha y^2 + \beta xy + \gamma x^2 + \delta y + \varepsilon x + \zeta = 0\] 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 \((0,0),\) \((1,1),\) \((2,2),\) \((3,3)\) and \((4,4)\) give rise to a system only of rank 3, and there is no unique solution.

The equation \[\alpha y^2 - (\alpha + \gamma)xy + \gamma x^2 +\delta y - \delta x = 0\] can be factored as

\[(y-x)(\alpha y - \gamma x + \delta) = 0.\]

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 \[(y-x)(\alpha y - \gamma x + \delta) = 0\] becomes \[(y-x)(y-mx-b)=0.\]

We note that if \(\alpha = 0\) in the factored equation \[(y-x)(\alpha y - \gamma x + \delta) = 0,\] then the second equation is of a vertical line. In this case, the two coefficients in question are \(\alpha = 0,\) and the \(x\)-value \({\delta}/{\gamma}.\)