hw7solsmc - , b 2(mod 3). Note that a 2 2 = 0(mod 3) and b...

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Solutions to Homework 7 - Math 2000 The solutions to all exercises except 4.12 may be found in the back of the book. (# 4.12) Let a, b Z . Prove that if a 2 + 2 b 2 0(mod 3), then either a and b are congruent to 0 modulo 3 or neither is congruent to 0 modulo 3. Solution . Note that this may be divided into cases. Every integer is congruent to either 0,1, or 2 modulo 3. Therefore we have either a 0(mod 3) , a 1(mod 3) , or a 2(mod 3) . Likewise b 0(mod 3) , b 1(mod 3) , or b 2(mod 3) . Thus we have nine cases: Case 1: a 0(mod 3) , b 0(mod 3). Note that a 2 0 2 = 0(mod 3) and b 2 0 2 = 0(mod 3). Thus a 2 + 2 b 2 0 + 2 · 0(mod 3) or a 2 + 2 b 2 0(mod 3) Case 2: a 0(mod 3) , b 1(mod 3). Note that a 2 0 2 = 0(mod 3) and b 2 0 2 = 0(mod 3). Thus a 2 + 2 b 2 0 + 2 · 0(mod 3) or a 2 + 2 b 2 0(mod 3) Case 3: a 0(mod 3) , b 2(mod 3). Note that a 2 0 2 = 0(mod 3) and b 2 0 2 = 0(mod 3). Thus a 2 + 2 b 2 0 + 2 · 0(mod 3) or a 2 + 2 b 2 0(mod 3) Case 4: a 1(mod 3) , b 0(mod 3). Note that a 2 0 2 = 0(mod 3) and b 2 0 2 = 0(mod 3). Thus a 2 + 2 b 2 0 + 2 · 0(mod 3) or a 2 + 2 b 2 0(mod 3) Case 5: a 1(mod 3)
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Unformatted text preview: , b 2(mod 3). Note that a 2 2 = 0(mod 3) and b 2 2 = 0(mod 3). Thus a 2 + 2 b 2 0 + 2 0(mod 3) or a 2 + 2 b 2 0(mod 3) Case 6: a 1(mod 3) , b 2(mod 3). Note that a 2 2 = 0(mod 3) and b 2 2 = 0(mod 3). Thus a 2 + 2 b 2 0 + 2 0(mod 3) 1 or a 2 + 2 b 2 0(mod 3) Case 7: a 2(mod 3) , b 0(mod 3). Note that a 2 2 = 0(mod 3) and b 2 2 = 0(mod 3). Thus a 2 + 2 b 2 0 + 2 0(mod 3) or a 2 + 2 b 2 0(mod 3) Case 8: a 2(mod 3) , b 1(mod 3). Note that a 2 2 = 0(mod 3) and b 2 2 = 0(mod 3). Thus a 2 + 2 b 2 0 + 2 0(mod 3) or a 2 + 2 b 2 0(mod 3) Case 9: a 2(mod 3) , b 2(mod 3). Note that a 2 2 = 0(mod 3) and b 2 2 = 0(mod 3). Thus a 2 + 2 b 2 0 + 2 0(mod 3) or a 2 + 2 b 2 0(mod 3) 2...
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hw7solsmc - , b 2(mod 3). Note that a 2 2 = 0(mod 3) and b...

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