Case 2 a is the number of pairs is odd number b, c, is an odd number of. Given a = 2m, b = 2k, c = 2n + 1 + 1 when n, m, k is a member of I +.A2 + b2 = is that c2. 4m2+4n2+4n+1 = 4k2+4k+1 k2-m2-n2+k-n = 0 (k-n)(k+n+1) = m2 Because n is a member of the I +, so m =. Select the value k and n that makes m is a member of I +.K = 2, n = 1; = 2 as a member of I +.For example, a triangle that has the cathetus is 8, 15, and hypotenuse equal to 5.
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