Question

please check my answer on pythagorean triples
http://gyazo.com/89185775b57fd55f8ac22bb00d4cf74d

Answer 1

It should be 45 and 53

Answer 2

hint, let 28 = 2xy

Answer 3

xy = 14

Answer 4

right, now you need pairs of factors of 14

Answer 5

how many ways can you factor 14, using two numbers

Answer 6

how do i do that

Answer 7

14 = 2x7

Answer 8

thats one way :)

Answer 9

what about 14 and 1 , 1 x 14 = 14, yes?

Answer 10

so total 3 ways

Answer 11

two ways

Answer 12

1x14 = 14
2 x 7 = 14

Answer 13

1x14
14x1

2x7
7x2

Answer 14

the order does not count

Answer 15

so only two options will be correct

Answer 16

we now can set x and y as 1,14 or 2,7

Answer 17

note that only one order will work, so as to avoid a negative number

Answer 18

ok

Answer 19

x=2
y=7
hypotenuse = 7^2+2^2=53

x=14
y=1
hypotenuse = 14^2+1^2=197

my answer is correct ?

Answer 20

http://www.math.rutgers.edu/~erowland/tripleslist-long.html

Answer 21

(x^2 - y^2, 2xy, x^2 + y^2)

x = 1 , y = 14
this does not work because 1^2 - 14^2 is negative

x = 14, y = 1
(14^2 - 1^2, 2*14*1, 14^2 + 1^2) = ( 195, 28, 197)

x= 2 , y = 7
this does not work because 2^2 - 7^2 is negative

x = 7, y = 2
(7^2 - 2^2, 2*7*2, 7^2 + 2^2) = ( 45, 28, 53)

Answer 22

pythagorean triples should be positive