Hello,
Historically, the following equations seem to be known to find the Pythagorean Numbers:
Equation(1): X=2a+1, Y=2a^2+2a, Z=2a^2+2a+1
Equation(2): X=4a, Y=4a^2-1, Z=4a^2+1
Equation(3): X=a^2-b^2, Y=2ab, Z=a^2+b^2
Recently, however, a Korean student insists in his thesis that no single eqation outof the above cannot give all the Pythagorean Numbers(PN's), but he has found himself the new equation from which all the PN's can be found as follows.
New Equation:
AB = 2 k^2, k= 1,2,...
A and B are devisers then,
all of the PN's can be found from the following equation:
X=(2AB)^(1/2)+A, Y=(2AB)^(1/2)+B, Z=(2AB)^(1/2)+A+B
Will someone comment who has interests on the subject above ?
Stefano/060808
Historically, the following equations seem to be known to find the Pythagorean Numbers:
Equation(1): X=2a+1, Y=2a^2+2a, Z=2a^2+2a+1
Equation(2): X=4a, Y=4a^2-1, Z=4a^2+1
Equation(3): X=a^2-b^2, Y=2ab, Z=a^2+b^2
Recently, however, a Korean student insists in his thesis that no single eqation outof the above cannot give all the Pythagorean Numbers(PN's), but he has found himself the new equation from which all the PN's can be found as follows.
New Equation:
AB = 2 k^2, k= 1,2,...
A and B are devisers then,
all of the PN's can be found from the following equation:
X=(2AB)^(1/2)+A, Y=(2AB)^(1/2)+B, Z=(2AB)^(1/2)+A+B
Will someone comment who has interests on the subject above ?
Stefano/060808