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Now let's suppose we have two cubic equations that are not multiples of one another. By Bézout's Theorem, it's possible for their curves to intersect in 9 points. Maclaurin, Cramer and Euler were all perplexed by this situation, because they knew the counting argument that shows that 9 points should generally give a unique curve of degree 3. The situation is even worse for \(n > 3.\) For example, two curves of degree \(n=4\) can intersect in \(n^2=16\) places and yet 14 points should suffice to determine such a curve uniquely. Here's what Maclaurin wrote about it (see [Maclaurin 1720, p. 137] or [Scott 1898, p. 260]):

A line of order \(n\) may intersect another of the same order in \(n^2\) points. Accordingly, when two lines of order \(n\) pass through \(n^2\) points, the given points, of which the number is \(\frac{1}{2}(n^2+3n),\) are not sufficient for determining the line of order \(n\) whose curve would be the unique curve that passes through the given points. … Thus, if nine points don't sufficiently determine a line of the third order, … nevertheless, 10 are more than enough for determining a line of the third order.

Maclaurin apparently could not resolve the paradox, but it did not cause him to abandon the proposition that \(\frac{1}{2}(n^2+3n)\) points determine a curve of order \(n.\) This was not the case for Braikenridge. Because he knew that two curves of degree \(n\) could intersect in \(n^2\) points, he concluded that the number of points needed to determine such a curve uniquely had to be greater than \(n^2.\) In [Braikenridge 1735], he argued that the number of points needed was \(n^2+1.\)

Cramer read Braikenridge's article and brought it to Euler's attention in a letter of 30 September 1744, the letter that preceded [Euler 1744b] in their correspondence. Cramer did not mention Maclaurin's argument in his letter to Euler, but Euler replied as follows [Euler 1744b, p. 2]:

I have seen that Mr. Maclaurin already had the same doubt concerning the number of points which determine curves of a given order: he says that to determine a line of the third order, the number of nine points may be too small, yet still the number of ten is too great, which in my opinion is an overt contradiction. The aforementioned Braikenridge is also absolutely mistaken in holding that a line of order \(n\) may be described by \(n^2+1\) points and it is a disputed truth, as you have very well remarked, that this number is but \(\frac{nn+3n}{2}.\) Furthermore, one may not doubt that two curved lines, one of which is of order \(m\) and the other of order \(n,\) may intersect in \(mn\) points, though you will be the first to have given a perfect proof of this truth, for I freely admit that my proof is all but complete.

It is interesting to note the status of Bézout's Theorem in 1744. Cramer had told Euler that he had a proof of it, which would soon appear in his book [Cramer 1750]. It would actually take six years for the book to be published. Meanwhile, Euler figured that he had just proved it himself, more or less, so that Cramer would still be the first to publish. It turned out that neither Euler's proof (in [Euler 1750b]) nor Cramer's were considered complete, so the theorem remained an open conjecture for another generation to tackle.

Euler continued in his letter to Cramer as follows [Euler 1744b, p. 2]:

At first, all of these reflections only served to bring to my attention the difficulties of the case, which you were so good as to propose to me. However, I finally found the solution to this doubt, with which I hope you will be satisfied. I say, then, that although it is indeed true that a line of order \(n\) be determined by \(\frac{nn+3n}{2}\) points, this rule is nevertheless subject to certain exceptions. For although the general equation of lines of order \(n\) has \(\frac{nn+3n}{2}\) coefficients to be determined, it may happen that such a number of equations, which we draw from the same number of given points, is not sufficient for this effect: this is evident, when two or several of these equations become identical. In such a case, one finds after having reduced the matter to the determination of the final coefficient, the value of this is expressed by a fraction, whose numerator and denominator both become =0. I conceive therefore, that this inconvenience will take place when the nine points, which ought to determine a line of the 3rd order, are disposed such that two curved lines of this order may be drawn through them. In this case, the nine given points, since they contain two identical equations, are worth but 8, and we may then add the tenth point in order to render the problem determined.

Euler did not have the language of linear independence and rank, but his resolution of the paradox can be expressed as follows: if we choose 9 points on a cubic curve that happen to be the same 9 points where that curve intersects a different cubic curve, then that collection of points is not suitably "generic" and it gives rise to a system of 9 equations in 10 unknowns with a rank of only 8. In this case, there will be two parameters in the solution space, so the solutions, which give the coefficients of the general cubic equation \[\alpha x^3 + \beta x^2y + \gamma xy^2 + \delta y^3 + \varepsilon x^2 + \zeta xy + \eta y^2 + \theta x + \iota y + \kappa = 0,\] will not all be scalar multiples of one another. When this happens, there really will be infinitely many distinct cubic curves passing through the 9 given points.

Working in the 18th century, Euler had neither the definitions nor the results of linear algebra at his disposal, so a proof of this assertion was beyond his means (see [Dorier 1995] for more on the state of vector space theory in the 1700s). In his letter to Cramer, he could not even give a concrete example of a set of 9 points with the property that infinitely many cubic curves pass through them. Instead, he had to be content with arguing by analogy with the case where 5 points do not determine a unique conic when 4 or more of them lie in a straight line. However, it would not be long before Euler came up with a beautiful example to illustrate the correctness of his claim.

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