form, this becomes: is odd (Pythagoras, c. 540 BC): is even (Plato, c. 380 BC): It can be shown that all Pythagorean triples can be obtained, with appropriate rescaling, from the basic Platonic sequence (, and ) by allowing to take non-integer rational values. If is replaced with the fraction in the sequence, the result is equal to the 'standard' triple generator (2mn, ,) after rescaling. It follows that every triple has a corresponding rational value which can be used to generate a similar triangle (one with the same three angles and with sides in the same proportions as the original). For example, the Platonic equivalent of is generated by as . The Platonic sequence itself can be derived by following the steps for 'splitting the square' described in Diophantus II.VIII. The Jacobi–Madden equation The equation, is equivalent to the special Pythagorean triple, There is an infinite number of solutions to this equation as solving for the variables involves an elliptic curve. Small ones are, Equal sums of two squares One way to generate solutions to is to parametrize a, b, c, d in terms of integers m, n, p, q as follows: Equal sums of two fourth powers Given two sets of Pythagorean triples, the problem of finding equal products of a non-hypotenuse side and the hypotenuse, is easily seen to be equivalent to the equation, and was first solved by Euler as . Since he showed this is a rational point in an elliptic curve, then there is an infinite number of solutions. In fact, he also found a 7th degree polynomial parameterization. Descartes' Circle Theorem For the case of Descartes' circle theorem where all variables are squares, Euler showed this is equivalent to three simultaneous Pythagorean triples, There is also an infinite number of solutions, and for the special case when , then the equation simplifies to, with small solutions as and can be solved as binary quadratic forms. Almost-isosceles Pythagorean triples No Pythagorean triples are isosceles, because the ratio of the hypotenuse to either other side is , but cannot be expressed as the ratio of 2 integers. There are, however, right-angled triangles with integral sides for which the lengths of the non-hypotenuse sides differ by one, such as, and an infinite number of others. They can be completely parameterized as, where {x, y} are the solutions to the Pell equation . If , , are