There is another way to construct all the Pythagorean triples that uses geometry instead of algebraic formulas. One starts with the circle x² + y² = 1 inscribed in a square, as shown in the figure on the next page. A line joining the upper right corner Po of the square to the midpoint P₁ of its lower edge intersects the circle in a point Q₁. This is in fact the rational point (4/5, 3/5). Next, make Q₁ the corner of a rectangle inscribed in the circle and join the other three corners P2, P3, and P4 of this rectangle to Po by lines intersecting the circle in points Q2, Q3, and Q4. These are also rational points on the sphere, with coordinates as shown. Now the process is to be repeated: Make each of Q2, Q3, and Q4 the corner of a rectangle inscribed in the circle, and join each of the other three corners of these rectangles to Po, producing lines that intersect the circle in new rational points, and so on. We could also have taken P₁ to be the midpoint of the left edge of the outer square and done the construction from there to produce more rational points, but these would just be the images of the ones already constructed under reflection across the line x = y, interchanging x and y coordinates. It is a fact that all the rational points on the circle can be constructed this way, and each one occurs only once. P P Po Q₁ = (.) Q=(.) Q-(%) Q-(f.) The technique we used to find the rational points on the circle x² + y² = 1 can also be used to find all the rational points on other quadratic curves Ax² + By² = C with integer or rational coefficients A, B, and C, provided that one can find a single rational point (x,y) on the curve to start the process, by considering lines through this point. For example, the circle x²+ y² = 2 contains the rational points (+1, +1) and we can use one of these as an initial point. Taking the point (1,1), we would consider lines y-1=m(x-1) of slope m passing through this point. Solving this equation for y and plugging into the equation x²+y2-2 we could then solve the resulting quadratic equation for x in terms of the slope m. Finally, the equation y-1=m(x-1) could be used to get y in terms of m as well. What makes the point (x,y) a rational point whenever the slope m is rational is the general fact that when a quadratic equation ax2 + bx+c=0 with rational coefficients has one rational root, then its other root must be rational as well. In the present case the quadratic equation has the root x-1 since we know the line and the circle intersect at (1,1). Hence the other root must also be rational. In practice, what this means is that the formulas for x and y in terms of m can be simplified until they each become the quotient of two polynomials in m. If instead of x²+y2-2 we consider the circle x²+y2-3 then there aren't any obvious rational points. And in fact this circle contains no rational points at all. For if there were a rational point, this would yield a solution of the equation a² + b² = 3c² by integers a, b, and c. We can assume a, b, and c have no common factor. Then a and b can't both be even, otherwise the left side of the equation would be even, forcing c to be even, so a, b, and c would have a common factor of 2. To complete the argument we look at the equation modulo 4. (This means that we consider the remainders obtained after division by 4.) The square of an even number has the form (2n)²-4n², which is 0 modulo 4, while the square of an odd number has the form (2n+1)24n2+4n+ 1, which is 1 modulo 4. Thus, modulo 4, the left side of the equation is either 0+1, 1+0, or 1+1 since a and b are not both even. So the left side is either 1 or 2 modulo 4. However, the right side is either 3-0 or 3-1 modulo 4. We conclude that there can be no integer solutions of a² + b² = 3c².