Solution to "Rational Right Angle Triangles with Equal Areas"

Right angled triangles with sides a and b and hypotenuse c obey Pythagoras' Theorem

a2 + b2 = c2

Suppose a, b and c are integers with no common factor. Then both a and b cannot be even, since, if they were, c would be even and 2 would be a common factor. Again, both a and b cannot be odd, since any odd square has form 4*n+1 and the sum of 2 odd numbers has form 8*n+2. This is divisible by 2 but not 4, and thus cannot be a square.
Thus, one side must be even and the other odd and c must odd. Call the even one b.
then b2 = c2 -a2 = (c-a)*(c+a)
c-a and c+a are unequal factors of a square number and hence must both be squares. Furthermore, they are both even. Thus let

4*u2 = c+a and 4*v2 = c-a.


b = 2*u*v and a = u2 - v2 and c = u2 + v2

Similar arguments show that u > v, u and v have no common factor, and they cannot both be odd.

The area of a right angled triangle is a*b/2 = u*v*(u2-v2)
If the sides have a common factor k then the area is

k2*a*b/2 = k2*u*v*(u2-v2)

My program used this formula to generate the areas of all integer-sided right angled triangles by letting k and u be any integers and v be any integer less than u with no common factor and u and v not both odd. It used values of u from 2 to 100 and then multiplied by k2 while the area was less than 1 million. Each triple of (area, a, b) was placed in a file and, when complete, the file was sorted in order of increasing area. This brought together triangles with the same area and a simple search program found the first group of 4 of the same area.
Their sides are


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