# When Nine Points Are Worth But Eight: Euler’s Resolution of Cramer’s Paradox - Higher Order Equations

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Now that we understand the relationship between a second order equation and the various kinds of conic sections, let's turn our attention to equations of higher order. A polynomial equation in two variables is an equation of the form $p(x,y)=0,$ where $$p(x,y)$$ is a polynomial. The terms of $$p(x,y)$$ have the form $$c x^i y^j,$$ where $$c$$ is a constant coefficient and $$i$$ and $$j$$ are non-negative integers.  We assume that $$p(x,y)$$ has been simplified so that there is only one such term for any particular pair $$i$$ and $$j.$$ The degree of the term $$c x^i y^j$$ is $$i+j.$$  Clearly, there can only be one term of degree 0, two terms of degree 1 and, in general, $$n+1$$ terms of degree $$n.$$

The degree of a polynomial equation of the form $$p(x,y)=0$$ is the maximum of the degrees of the terms of $$p(x,y).$$ Therefore, the general form of a polynomial equation of degree one (a linear equation) is

$A x + B y + C = 0.$

The general form of a polynomial equation of degree two (a quadratic equation) was written by Euler as

$\alpha y^2 + \beta xy + \gamma x^2 + \delta y + \varepsilon x + \zeta = 0.$

The general form of a polynomial equation of degree $$n$$ is

$\sum^{n}_{k=0} \sum^{k}_{i=0} \alpha_{k,i} x^i y^{k-i} = 0,$

where the $$\alpha_{k,i}$$ are constant coefficients and there is at least one $$i_0$$ with $$0 \le i_0 \le n$$ satisfying $$\alpha_{n,i_0} \ne 0.$$  Using the familiar formula for the sum of the first $$N$$ integers, the number of coefficients in a polynomial equation of degree $$n$$ is the sum $1 + 2 + 3 + \ldots + n + (n+1) = \frac{(n+1)(n+2)}{2} = \frac{n^2+3n}{2} + 1.$

A curve of degree $$n$$ (called a "line" of degree $$n$$ by Euler and most other 18th century authors) is the graph of the solution set of a polynomial equation of degree $$n.$$  An equation of the form $$p(x,y)=0$$ may be multiplied by an arbitrary non-zero constant without changing the set of pairs $$(x,y)$$ that satisfy it. Therefore, in order to determine its solution set, it's only necessary to specify the ratios among the coefficients of $$p(x,y),$$ not the coefficients themselves. The number of such ratios, denoted $$\varphi_n,$$ is one less than the number of coefficients; that is $\varphi_n = \frac{n^2+3n}{2}.$