Question

How many ordered triples $(x,y,z)$ of integers are there such that $\sqrt{x^2 + y^2 + z^2} = 7$? Does the question have a geometric interpretation? Please help! Thanks!

Answer 1

First square both sides.

Answer 2

x^2 + y^2 + z^2 =7^2

Answer 3

then we will end up with a sphere with radius 7

Answer 4

so there are many triples. Is it correct?

Answer 5

You will end up with (0,0,7) ( 0, 7, 0) and ( 7, 0 , 0 ) as solutions

Answer 6

( 0, 0 , -7 ) ( 0,0,7)
(0,7 , 0) ( 0 , -7 , 0 )
(7,0,0) , (-7, 0, 0)

Answer 7

Two of the principal solutions are 0,0,7 and 6,3,2 , then you just rearrange them and then you take their negatives.

Answer 8

This leaves you with 54 integer solutions

Answer 9

I got it. Thank you!!

Answer 10

No problem!

Answer 11

@jchick how do you know there aren't more?

Answer 12

we know that (6,3,2) is a solution, and we can rearrange it , since they are symmetric
x^2 + y^2 + z^2 = 7^2
6^2 + 3^2 + 2^2 = 7^2
3^2 + 6^2 + 2^2 = 7^2

Answer 13

each rearrangement of 6,3,2 is another solution
x = 6 y = 3 z = 2
x = 6 , y = 2 , z = 3

Answer 14

6,3,2
6,2,3

3, 6,2
3,2,6

2,6,3
2,3,6

Answer 15

(0,0,7) , there are three ways to rerarrange that.
(0,7,0)
(7,0,0) , and times 2 since there is one neg/pos each

so
3!*2*2*2 + 3 * 2 is the number of integer solutions

Answer 16

@Kainui