This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 isnt 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 ,...
View
Full
Document
 '07
 LOZANOROBLEDO
 Number Theory, Formulas

Click to edit the document details