anonymous
  • anonymous
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!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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jchick
  • jchick
First square both sides.
jchick
  • jchick
x^2 + y^2 + z^2 =7^2
jchick
  • jchick
then we will end up with a sphere with radius 7

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jchick
  • jchick
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anonymous
  • anonymous
so there are many triples. Is it correct?
jchick
  • jchick
You will end up with (0,0,7) ( 0, 7, 0) and ( 7, 0 , 0 ) as solutions
jchick
  • jchick
( 0, 0 , -7 ) ( 0,0,7) (0,7 , 0) ( 0 , -7 , 0 ) (7,0,0) , (-7, 0, 0)
jchick
  • jchick
Two of the principal solutions are 0,0,7 and 6,3,2 , then you just rearrange them and then you take their negatives.
jchick
  • jchick
This leaves you with 54 integer solutions
anonymous
  • anonymous
I got it. Thank you!!
jchick
  • jchick
No problem!
Kainui
  • Kainui
@jchick how do you know there aren't more?
jchick
  • jchick
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
jchick
  • jchick
each rearrangement of 6,3,2 is another solution x = 6 y = 3 z = 2 x = 6 , y = 2 , z = 3
jchick
  • jchick
6,3,2 6,2,3 3, 6,2 3,2,6 2,6,3 2,3,6
jchick
  • jchick
(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
jchick
  • jchick
@Kainui

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