Part 1: Derivation of a New Procedure for Pythagorean Triples

The following is the derivation of a new procedure for generating Pythagorean triples – i.e., solutions of the Pythagorean equation that have integral values:

Step 1:

        x² + (x + a)² = (x + b)² 

        x² + x² + 2ax + a² = x² + 2bx + b² 

        x² + 2ax – 2bx + a² – b² = 0 

        x² + 2x(a – b) + (a² – b²) = 0 

        x² + 2x(a – b) + (a – b)² – (a – b)² + (a² – b²) = 0 

        [x + (a – b)]² – (a – b)² + (a² – b²) = 0 

        [x + (a – b)]² – a² + 2ab – b² + a² – b² = 0 

        [x + (a – b)]² + 2b(a – b) = 0 

        [x + (a – b)]² – 2b(b – a) = 0  

[x + (a – b)]² = 2b(b-a) 

        x + (a – b) = [2b(b – a)]^1/2 

        x = [2b(b – a)]^1/2 – (a – b) 

(1)    x = [2b(b – a)]^1/2 + (b – a) 

Step 2: 

Since x, x + a, and x + b are all integers, a and b must be integers as well. Therefore, a + b must also be an integer, and there must be some rational number c = x/(a + b), or: 

(2) x = c(a + b) 

Equating the right sides of (1) and (2), it follows that:

        c(a + b) = [2b(b – a)]^1/2 + (b – a)

        c(a + b) – (b – a) = [2b(b – a)]^1/2

        c²(a + b)² – 2c(a + b)(b – a) + (b – a)² = 2b² – 2ab

        a²c² + 2abc² + b²c² + 2a²c – 2b²c + b² – 2ab + a² = 2b² – 2ab

        a²c² + 2a2c – 2b2c + 2abc² + b²c² = b² – a²

        c²(a² + 2ab + b²) + (2c + 1)(a² – b²) = 0

        c²(a + b) + (2c + 1)(a – b) = 0

        ac² + bc² + 2ac – 2bc + a – b = 0

        (a/b)c² + c² + 2(a/b)c – 2c – a/b – 1 = 0

        (a/b)(c² + 2c + 1) = –c² + 2c + 1

        (a/b)(c + 1)² = – (c² – 2c – 1)

        (a/b)(c + 1)² = – (c² – 2c + 1 – 2)

        (a/b)(c + 1)² = 2 – (c² – 2c + 1)

        (a/b)(c + 1)² = 2 – (c – 1)²

(3)    a/b = 2 – (c – 1)²

                (c + 1)²

Step 3:

Select any rational number c, where c < -1 or 0 < c. Solve (3) for a/b. Set the numerator of a/b equal to a and the denominator equal to b. Solve (1) or (2) for x, which yields x + a and x + b. This is the procedure in its entirety.

Note: taking – 1 < c < 0 does not yield Pythagorean triples. At least one of x, x + a, or x + b is 0 or negative.

Note: for 0 < c </ (1 + 21/2, x < x + a < x + b, and for c < -1 and (1 + 21/2) < c, x + a < x < x + b.

Note: c must be rational. Assume the contrary. It follows that x/c = (a + b) is irrational. From this, it follows that a or b or both are irrational. From this, it follows that (x + a) or (x + b) or both are irrational.

Note: by substitution into (2), the following limits obtain:

When c à infinity, the limit of x/(x + b) is 4/5, the limit of (x + a)/(x + b) is 3/5, and the limit of x/(x + a) is 4/3.

When c à 0, the limit of (x + a)/(x + b) is 1, the limit of x/(x + a) is 0, and the limit of x/(x + b) is 0.

When c = 2^1/2 + 1, a = 0, (x + a)/x = 1, (x + b)/x = 2^1/2, and (x + b)/(x + a) = 2^1/2. Thus, there can be no isosceles right triangle with all sides of integral length.

Part 2: Comparison of the Author’s Method with the Standard Procedure for Generating Pythagorean Triples

In Excursions in Number Theory (New York: Dover, 1966, pp. 66-67), Ogilvy and Anderson give this derivation of the standard method of generating Pythagorean triples, which dates back to ancient Greek and Babylonian times (c. 500 BC):

To guarantee that we get new triangles [as opposed to linear multiple ones] we can consider only the primitive solutions, meaning x, y, z having no factor in common…x and y cannot both be odd [according to congruence theory]. On the other hand, x and y cannot both be even, for then z would be even and the solution would not be primitive.

Suppose, then, that x is odd and y is even. This means that we can write x² + 4u² = z², with no two of x, u, z having a common factor. Therefore, 4u² = z² – x² = (z + x)(z – x). But x and z are odd. Hence, both (z + x) and (z – x) are even, say z + x = 2s, z – x = 2r. That is, 4u² = 2s2r or u² = rs. Now the two equations in z and x yield, on addition, z = r + s, and on subtraction, x = s – r. If r and s had a common factor, z and x would have it also. But z and x are relatively prime; therefore so are r and s. As a consequence, the equation u2 = rs requires that r and s are each perfects squares (neither can pick up a “matching” factor from the other) say, r = n² and s = m². Then u = mn, and we have finally:

        x = m² – n²

        y = 2mn

        z = m² + n²

…[W]hat are the restrictions on m and n? First, m > n, so that x will be positive. Second, m, n must have no common factor, or it would be shared (twice) by x, y, and z. Third, m and n must not both be odd, for then x and z would share the factor 2.

There is a simple relationship between the “inputs” of the traditional method of generating Pythagorean triples and the author’s method presented above. The derivation of the “transformation equations” (i.e., the equations that link the two methods) is as follows:

Given x odd and (m, n) restricted as above, a = 2mn – x = 2mn – m² + n², and b = m² + n² – x = 2n². So, since x = c(a + b), m² – n² = c(2mn – m² + n² + 2n²), and c = (m – n)/(3n – m). If x is even, c = (g + h)/(g – h) = n/(m – 2n), so that c = n/(m – 2n).

Given x odd and rational number c = g/h, where 0 < c or c < -1, g/n = (m – n)/(3n – m), mh – nh = 3ng – mg, m(h + g) = n(3g + h), and m/n = (3g + h)/(g + h). Set m = 3g + h and n = g + h. If x is even, m/n = (3[g + h] + [g – h]/([g + h] + [g – h]) = (4g + 2h)/2g, so that m/n = (2g + h)/g.