Generating the Measures of n-Balls

s == ( a 1 + a 2 )( a 2 + a 3 )( a 3 + a 1), 8 == s( a 1/( a 1 + a 2 ) + a 2/( a 2 + a 3 ) + a 3/( a 3 + a 1 ) ). It can be shown that (8js)2 + (8js) + djs 2 = 0 and (8/s) E K as in the case char K * 2. Apply Theorem 1 again. We find that both u3 jv and djs = (u3 jv ) + 1 belong to {w 2 + w E K: w E K}. Thus the equation x 2 + x + 1 = 0 has a solution in K. This finishes the proof. The reader may consult [1] for an account of the characteristic-two discriminant.