HW solutions 1

# HW solutions 1 - Math 3320 Problem Set 1 Solutions 1 1(a...

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Unformatted text preview: Math 3320 Problem Set 1 Solutions 1 1. (a) Make a list of the 16 primitive Pythagorean triples (a,b,c) with c ≤ 100, regarding (a,b,c) and (b,a,c) as the same triple. Solution : Plugging in values for p and q of opposite parity in the formulas x = 2 pq , y = p 2- q 2 , z = p 2 + q 2 gives the triples ( 4 , 3 , 5 ) , ( 12 , 5 , 13 ) , ( 8 , 15 , 17 ) , ( 24 , 7 , 25 ) , ( 20 , 21 , 29 ) , ( 40 , 9 , 41 ) , ( 12 , 35 , 37 ) , ( 60 , 9 , 61 ) , ( 28 , 45 , 53 ) , ( 56 , 33 , 65 ) , ( 84 , 13 , 85 ) , ( 16 , 63 , 65 ) , ( 48 , 55 , 73 ) , ( 80 , 39 , 89 ) , ( 36 , 77 , 85 ) , ( 72 , 65 , 97 ) . (b) How many more would there be if we allowed nonprimitive triples? Solution : For the primitive triple ( 4 , 3 , 5 ) we get 19 nonprimitive triples with c ≤ 100 by multiplying ( 4 , 3 , 5 ) by 2 , 3 , 4 , ··· , 20. Similarly, ( 12 , 5 , 13 ) gives 6 more, ( 8 , 15 , 17 ) gives 4 more, ( 24 , 7 , 25 ) gives 3 more, ( 20 , 21 , 29 ) gives 2 more, and both ( 40 , 9 , 41 ) and (( 12 , 35 , 37 ) give 1 more, for a grand total of 36 more. (c) How many triples (primitive or not) are there with c = 65? Solution : The two in part (a) are ( 56 , 33 , 65 ) and ( 16 , 63 , 65 ) . There are two more that are not primitive, obtained by multiplying ( 4 , 3 , 5 ) by 13 and ( 12 , 5 , 13 ) by 5, so the total number is 4. 2. Show that there are no Pythagorean triples (a,b,c) with a being a positive integer multiple of b , or vice versa. Solution : If a 2 + b 2 = c 2 and a = nb then we get n 2 b 2 + b 2 = c 2 . The left side of this equation is divisible by b 2 , so c 2 is divisible by b 2 , which implies that c is divisible by b . The equation can then be written as n 2 + 1 = ( c b ) 2 = m 2 . Thus the difference between the squares n 2 and m 2 is 1. The only time the difference between two squares is 1 is for the squares 0 and 1, so n = 0 and m = 1. But if n = 0 then the equation a = nb gives a = 0, which isn’t allowed in a Pythagorean triple. 3. (a) Find all the positive integer solutions of x 2- y 2 = 512 by factoring x 2- y 2 as (x + y)(x- y) and considering the possible factorizations of 512. Solution : The equation (x + y)(x- y) = n can be solved by factoring n as n = pq to obtain two equations x + y = p and x- y = q . Solving for x and y , we get x = (p + q)/ 2 and y = (p- q)/ 2. In order for x and y to be positive integers, p and q must have the same parity (both even or both odd), and we must have p > q . For n = 512 = 2 9 the only possibilities are then (p,q) = ( 256 , 2 ), ( 128 , 4 ), ( 64 , 8 ), ( 32 , 16 ) with corresponding solutions (x,y) = ( 129 , 127 ), ( 66 , 62 ), ( 36 ,...
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HW solutions 1 - Math 3320 Problem Set 1 Solutions 1 1(a...

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