Question 97137
Obviously the pythagorus triple: 6,8,10
One leg = x
Other leg = x + 2
a^2 + b^2 = c^2
x^2 + (x + 2)^2 = 10^2
x^2 + x^2 + 4x + 4 = 100
2x^2 + 4x - 96 = 0
x^2 + 2x - 48 = 0
(x - 6)(x + 8) = 0
x = 6 and -8
Other leg: 8 or -6
Obviously we cannot accept negative values ...
6 and 8