Rational Triangles

A rational triangle can be defined as one having all sides with rational length.

Rational Right Triangles - Early Solutions
In Śulba solution for the equation $$x^2+y^2=z^2$$---(1) is available. Baudhāyana (c. 800 B.C.), Āpastamba and Kātyāyana (c. 500 B.C.) gave a method for the transformation of a rectangle into a square, which is the equivalent of the algebraical identity.

$${\displaystyle mn = \left (m- \frac{m-n}{2} \right)^2 - \left (\frac{m-n}{2} \right)^2 } $$

where m, n are any two arbitrary numbers. Thus we get

$${\displaystyle =(\sqrt{mn})^2+\left ( \frac{m-n}{2} \right )^2= \left ( \frac{m+n}{2} \right )^2}$$

substituting p2,q2 for m, n respectively in order to eliminate the irrational quantities, we get

$${\displaystyle =p^2q^2+\left ( \frac{p^2-q^2}{2} \right )^2= \left ( \frac{p^2+q^2}{2} \right )^2}$$

which gives the rational solution of (1).

Kātyāyana gives a very simple method for finding a square equal to the sum of a number of other squares of the same size which gives us with another solution of the rational right triangle.

Kātyāyana says: "As many squares (of equal size) as you wish to combine into one, the transverse line will be (equal to) one less than that; twice a side will be (equal to) one more than that; (thus) form (an isosceles) triangle. Its arrow (i.e., altitude) will do that."



For combining n squares of sides a each we form the isosceles triangle ABC such that $$AB=AC=\frac{(n+1)a}{2}$$ and $$BC=(n-1)a$$

Then $$AD^2=na^2$$ which gives the formula

$${\displaystyle =a^2(\sqrt{n})^2+a^2\left (\frac{n-1}{2} \right )^2= a^2\left ( \frac{n+1}{2} \right )^2}$$

put m2 for n in order to make the sides of the right angled triangle  without the radical, we have

$${\displaystyle =m^2a^2+a^2\left (\frac{m^2-1}{2} \right )^2= a^2\left ( \frac{m^2+1}{2} \right )^2}$$ which gives the rational solution of (1).