D = 15. (a) Show that we get four reduced quadratic forms AX2 +2BXY +CY2 and that the classnumber...

D = 15. (a) Show that we get four reduced quadratic forms AX2 +2BXY +CY2 and that the classnumber h(Q(v-15)) = 2. (b) Show that X2 + 15Y 2 properly represents 24 = 16 and 26 = 64, but not 25 = 32 and 27 = 128. (c) Now consider the case 2m+2 =2 8 = 256. Start with Q1(X,Y) = 256X2 + 78XY +6Y 2 and Q2(X,Y)= 256X2 +178XY +31Y 2. Show that Q1(X,Y) is equivalent to 4X2 +2XY +4 Y 2and that Q2(X,Y) is equivalent to X2 +15Y 2. Caution The reduction of Q1(X,Y) to 4X2+2XY+4Y 2 transforms (1,0) to the primitive solution (6,-7), which will cause a negative answer of our compliance test. Clearly, a transposition X ?? Y would resolve our problem ... (d) Finally, consider the case 2m+2 =2 9 = 512. Start with Q1(X,Y) = 512X2 + 78XY +3Y 2 and Q2(X,Y)= 512X2 +434XY +92Y 2. Show that Q1(X,Y) is equivalent to 3X2 +5Y 2and that Q2(X,Y) is equivalent to 2X2 +2XY +8Y 2. Thus D = 15 is not a CM discriminant for 27 = 128.