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HW solutions 10 - Math 3320 Problem Set 10 Solutions 1...

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Math 3320 Problem Set 10 Solutions 1 1. Determine the values of Δ for which there exists a quadratic form of discriminant Δ that represents 5, and also determine the discriminants Δ for which there does not exist a form representing 5. Solution : The criterion for the existence of a form of discriminant Δ representing n is that Δ is a square mod 4 n , so for n = 5 the condition is that Δ is a square mod 20. The squares mod 20 are 0 2 = 0, ( ± 1 ) 2 = 1, ( ± 2 ) 2 = 4, ( ± 3 ) 2 = 9, ( ± 4 ) 2 = 16, ( ± 5 ) 2 = 25 5, ( ± 6 ) 2 = 36 16, ( ± 7 ) 2 = 49 9, ( ± 8 ) 2 = 64 4, ( ± 9 ) 2 = 81 1, and 10 2 = 100 0. So the discriminants for which 5 is represented are the Δ 0 , 1 , 4 , 5 , 9 , 16 mod 20. Discriminants are numbers congruent to 0 or 1 mod 4, so mod 20 this means 0 , 1 , 4 , 5 , 8 , 9 , 12 , 13 , 16 , 17. Thus the discriminants for which 5 is not represented are the Δ 8 , 12 , 13 , 17 mod 20. 2. Verify that the statement of quadratic reciprocity is true for the following pairs of primes (p, q) : ( 3 , 5 ) , ( 3 , 7 ) , ( 3 , 13 ) , ( 5 , 13 ) , ( 7 , 11 ) , and ( 13 , 17 ) . Solution : First we can make a list of the nonzero squares mod p for the primes p that we’re interested in: p = 3 : 1 5 : 1 , 4 7 : 1 , 2 , 4 11 : 1 , 3 , 4 , 5 , 9 13 : 1 , 3 , 4 , 9 , 10 , 12 17 : 1 , 2 , 4 , 8 , 9 , 13 , 15 , 16 To simplify the notation, let us introduce the so-called Legendre symbol parenleftBig p q parenrightBig which is defined to be parenleftBig p q parenrightBig = + 1 if p is a square mod q and parenleftBig p q parenrightBig = - 1 if p is not a square mod q . Here p and q are assumed to be distinct odd primes. The law of quadratic reciprocity then says parenleftBig p q parenrightBig = parenleftBig q p parenrightBig unless p and q are both congruent to 3 mod 4, in which case parenleftBig p q parenrightBig = - parenleftBig q p parenrightBig . For the first pair (p, q) = ( 3 , 5 ) we see from our list above that 3 is not a square mod 5 and neither is 5 a square mod 3. Thus we have parenleftBig 3 5 parenrightBig = - 1 and parenleftBig 5 3 parenrightBig = -
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