# When Nine Points Are Worth But Eight: Euler’s Resolution of Cramer’s Paradox - Five in a Row

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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 ${\delta}/{\gamma}.$

Robert E. Bradley (Adelphi University) and Lee Stemkoski (Adelphi University), "When Nine Points Are Worth But Eight: Euler’s Resolution of Cramer’s Paradox - Five in a Row," Convergence (February 2014)