anonymous
  • anonymous
How many solutions which integers such that x^2 + y^2 + z^2 = 2016
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ganeshie8
  • ganeshie8
I think it is sufficient to consider solutions in positive integers first, then we can build the complete solutions by permutations.
ganeshie8
  • ganeshie8
Working mod 4 it is easy to see that each of x, y, z must be of form \(4k\). Otherwise left hand side is not divisible by 16. So the equation reduces to \[a^2+b^2+c^2 = 126\] where \(x = 4a\), \(y=4b\), \(z=4c\)
anonymous
  • anonymous
apparently gauss did it over 200 years ago but hmm

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ganeshie8
  • ganeshie8
related to quadratic reciprocity law in any ways ? maybe taking mod 7 or 3 might narrow it further idk... my attempt above looks very hacky..
anonymous
  • anonymous
actually, he did it for square free numbers, so I guess we're screwed
ganeshie8
  • ganeshie8
looks interesting http://gyazo.com/301173b00539b4577e227beafd0b4281
dan815
  • dan815
|dw:1436766690043:dw|
dan815
  • dan815
|dw:1436766915671:dw|
dan815
  • dan815
|dw:1436767224913:dw|
dan815
  • dan815
lets start by considering the total number of ordered pairs of inteegers in this arc
dan815
  • dan815
then we can see which one of them will produce an int for the z coordinate too
ganeshie8
  • ganeshie8
There is a nice formula for number of representations of \(n\) as sum of two squares though
ganeshie8
  • ganeshie8
\[x^2+y^2 = 2016\] has no solutions over integers
dan815
  • dan815
|dw:1436767447283:dw|
dan815
  • dan815
this will give us a max bound on the number of integer solutions atleast
ganeshie8
  • ganeshie8
we can discard 2016 and work on smaller number 126 because both equations must have same number of solutions
dan815
  • dan815
how come we can do that?
ganeshie8
  • ganeshie8
\(x^2+y^2+z^2=2016 = 16*126 \) consider this equation in mod 4 suppose \(x\equiv y\equiv z\equiv 0 \pmod{4}\) then \(x^2+y^2+z^2 = 16(a^2+b^2+c^2) = 2016 \implies a^2+b^2+c^2 = 126\)
dan815
  • dan815
but we dont have to be able to factor 16 from all the squares
ganeshie8
  • ganeshie8
left hand side wont be divisible by \(16\) for all other cases, so we don't need to wry about them
ganeshie8
  • ganeshie8
for example, \(x\equiv y\equiv z\equiv 1\pmod{4}\) makes left hand side \(\equiv 3 \pmod{16}\) but the right hand side is \(\equiv 0 \pmod{16}\) similarly you can show that all the remaining cases don't work
dan815
  • dan815
what if we are dealing with cases where x y and z are all different mod and end up adding to 0 mod 16
ganeshie8
  • ganeshie8
thats exactly what im saying, thats not possible. maybe consider another example : \(x\equiv 2, y\equiv 1,z\equiv 0 \pmod{4}\) then \[x^2\equiv 0,~y^2\equiv 1, ~z^2\equiv 0\] clearly left hand side is \(\equiv 1 \pmod{4}\) but the right hand side is \(\equiv 0\pmod{4}\)
dan815
  • dan815
okk i see what u are saying
dan815
  • dan815
okay one more question, dont we have to consider the behavior in other mods too?
ganeshie8
  • ganeshie8
Nope, any integer can be represented in any one of the form \(4k\pm a\) where \(0\le a\lt 4\). s we're done if we're done in mod 4
ganeshie8
  • ganeshie8
this is same as considering even/odd cases right
ganeshie8
  • ganeshie8
If we prove/disprove an integer equation for both odd and even cases, then we're done. working even/odd cases is same as mod2, we dont need to wry about other mods
dan815
  • dan815
ok isee
dan815
  • dan815
can we continue the process then
ganeshie8
  • ganeshie8
Any square number is either \(0\) or \(1\) in \(\mod 4\), and clearly \(X+Y+Z \not\equiv 0\) when at least one of the variables on left hand side is \(1\). so it must be the case that \(X\equiv Y\equiv Z\equiv 0\)
dan815
  • dan815
and factor 3^2 and solve in mode 3?
ganeshie8
  • ganeshie8
actually i was stuck with \[a^2+b^2+c^2 = 126\] mod 3 is not making it any simpler
dan815
  • dan815
11,2,1
dan815
  • dan815
the simplest one i can find
ganeshie8
  • ganeshie8
Finding each solution gives us 3! = 6 solutions by permuting the ordered pairs
dan815
  • dan815
the choices are pretty low to pick from, maybe they expect brute force from here
ganeshie8
  • ganeshie8
yeah
dan815
  • dan815
11 is the highest 10 is ruled out 100 + 26 x^2+y^2=26 has no inter solution 9 is ruled out too 81 + 35 x^2+y^2=35 no int solution
dan815
  • dan815
oh wait 25+1
dan815
  • dan815
10 is good too
dan815
  • dan815
nothing else
dan815
  • dan815
11,2,1 10,5,1
dan815
  • dan815
checking 9 81+45 45=6^2+3^2
dan815
  • dan815
9,6,3
ganeshie8
  • ganeshie8
thats all
dan815
  • dan815
11,2,1 10,5,1 9,6,3 and so on
dan815
  • dan815
theres gotta be a better way than brute force
ganeshie8
  • ganeshie8
there is a better way when the right hand side is squarefree, but 126 is not square free. so there is no known simple method
dan815
  • dan815
theres nothing with 8 and 7, so we are actually done here
dan815
  • dan815
since it will repeat
dan815
  • dan815
11,2,1 10,5,1 9,6,3 so 6*5*4 *3 right?
ganeshie8
  • ganeshie8
3*3! positive solitions 3*3!*2^3 integer solutions ?
dan815
  • dan815
i was thinking well its plus or minus 11,2,1 so we have 6 choices and 3 spots 6*5*4 so 6*5*4 for each triple ordered pair
dan815
  • dan815
oh oops ya you are right, if i do this ill allow -11 and 11
dan815
  • dan815
(6*4*2) * 3 = 3*3! * 2^3 oh this works too
dan815
  • dan815
|dw:1436769950610:dw|
ganeshie8
  • ganeshie8
yes 144 looks correct!
anonymous
  • anonymous
Yups the answer is 144. You all are best mathematiciants :)
anonymous
  • anonymous
Thank you very much for your answer guys

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