 Around the same time Descartes was writing the method, Pierre de Fermat had an insight. In his introduction to plain and solid loci, Fermat claimed, every time a final equation leaves two unknown quantities, one has a locus, the extremity of which it describes a straight line or a curve. Fermat went further. It is convenient to set these equations to let the two unknown quantities form a given angle, which we ordinarily take to be a right angle, and to set the position of the one at the extremity of the other. What Fermat is describing is what we would call the graph of an equation. Fermat went even further and identified equations with specific curves. For example, let nm be a line and bd constants take nz equal to a, and let zi equal to e be drawn at a specified angle to nm. If dA equals be, then i describes a straight line. Now it's worth pointing out here that for Fermat, b and d were constants, while a and e were what we would call variable quantities. In other words, Fermat is actually using Vietta's convention of having consonants be constants and vowels be variables. And Fermat proves that this actually is a straight line as follows. If dA equals be, then d over b is equal to e over a. So e over a have a constant ratio, since d and b are constants. So triangle nzi is similar to some specific triangle, and consequently, regardless of the location of i, the angle zni is fixed. Fermat goes on to note that the equation dr minus da equals be also describes a straight line. If he goes through Fermat's proof, we can gain some insight into his method. So rearranging this as a proportionality. So let mn equal r, some fixed quantity. Then mz is r minus a, and so zi to nz also has a constant ratio, and by almost the same argument as before, mn is similar to a specific triangle, etc. How about the conic sections? From the Greek Geometers, Fermat knew the locus relations for the conic sections. Let p be a constant, the abscissa x and ordinate y of a parabola satisfy the relationship p y equals x squared, and let q be a constant, the abscissa x and ordinate y of a hyperbola satisfy the relationship x y equals q. While the Greeks, especially after Apollonius, didn't require the abscissa and ordinate to be perpendicular, Fermat usually assumed they were. In that case, the vertex passes through the axis of a parabola, and the hyperbola's asymptotes are perpendicular. This allowed Fermat to identify other curves. For example, we might take completely at random the relationship where s, r, and d are constants, and a and e are the unknown lengths. For this case, Fermat proceeds as follows. Let nz equal to a, and zi equal to e. Now, we need a line corresponding to a minus s, which is nz minus some length. So make no equal to s, so a minus s is oz. We also need a line corresponding to r minus e, which is some fixed length minus zi. So make zp equal to r, so ip is r minus e. And we'll form the rectangle n dpz, so vp is oz is a minus s, and pi is r minus e. So if the rectangle on vp and pi has constant area d minus rs, then i will be on a hyperbola with asymptotes vp and vo. By the end of 1637, two Frenchmen, a lawyer and an ex-immersionary, had both applied algebraic methods to geometry. What could possibly go wrong? Well, nothing if only Fermat and Descartes were involved. However, Descartes had previously insulted a number of prominent French mathematicians who presented Fermat's work as a superior alternative to Descartes. Now remember, Fermat was a lawyer and not an academic, so he was probably unaware of the feud between Descartes and the French mathematicians. When presented with Fermat's results, Descartes felt Fermat's method of maxima and minima was flawed and, indirectly, challenged Fermat in a letter to Marin Marseille, dated January 18th, 1638. Descartes' language is rather inflammatory. He criticizes Fermat's method and says, moreover, his alleged rule is not as universal as he thinks it is, as it cannot be applied to any question of the least difficulty. To emphasize this point, Descartes posed the following problem, to find the tangent to a curve where the sum of the cubes on the ordnance is equal to the product of the ordnance and a given constant. In modern terms, Descartes is talking about the tangent to the curve we would write as x cubed plus y cubed equals pxy, where p is some constant. This curve is now known as the folium of Descartes. Now Marseille eventually conveyed the letter to Fermat, who solved the problem rather easily. We won't go into the details of the solution. Something has to be left for homework, but as it turns out, the problem is extremely challenging using the method Descartes described in the geometry, but very easy using Fermat's approach. Eventually Descartes and Fermat reconciled, or more properly, Descartes realized his feud was not with Fermat, and in the aftermath of this several things became clear. The Descartes-Fermat correspondence makes clear that the theory of equations was the basis for both the methods of Fermat and Descartes. In somewhat modernized terms, we can describe their basic approach as follows. For finding extreme values, in general, there are two values A and A plus E that give the same value of the objective function. We might view this as the fact that a horizontal line will cross the objective function at two points near an extreme value. But at the local extreme values, the two values are the same. Descartes' two roots become one. And so this means that we can find the extreme values by requiring our objective function equal to some maximum or minimum value have a repeated root. Similarly for finding tangents, in general, a non-tangent line will cross the curve at two points, but the two points will coincide if the line is tangent. So we can find the tangent line by requiring the curve and the tangent line have a double solution. At which point there is the following problem. How can we tell when an equation has a repeated root? And we'll take a look at that later.