A salutary or solitary point

Extract

Some three weeks into the academic year I presented, off the cuff, the following morsel to engage some lower-sixth formers who had polished off some routine derivatives all too quickly:

Sketch the curve whose equation is given by x³ + y³ + 3xy – 1 = 0 , and find the gradient at the point (–1 , –1).

Since this was near the end of a double period I thought no more of it. At the end of the next double period one student remarked that he had been having difficulty in sketching it. “It seems to be a straight line”, he said. I rashly suggested that he could well be mistaken, but promised to look at it. If my faith in the student had been greater I could have saved myself some labour searching for simple solutions of the cubic. I had noticed the solution (½ , ½) by the way. I began to have delusions of grandeur, imagining myself to be the natural successor of Diophantus, as I found the solutions (5,–4) and (–4,5).