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anonymous
 one year ago
How many solutions which integers such that x^2 + y^2 + z^2 = 2016
anonymous
 one year ago
How many solutions which integers such that x^2 + y^2 + z^2 = 2016

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I think it is sufficient to consider solutions in positive integers first, then we can build the complete solutions by permutations.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Working 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
 one year ago
Best ResponseYou've already chosen the best response.0apparently gauss did it over 200 years ago but hmm

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3related 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
 one year ago
Best ResponseYou've already chosen the best response.0actually, he did it for square free numbers, so I guess we're screwed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3looks interesting http://gyazo.com/301173b00539b4577e227beafd0b4281

dan815
 one year ago
Best ResponseYou've already chosen the best response.2lets start by considering the total number of ordered pairs of inteegers in this arc

dan815
 one year ago
Best ResponseYou've already chosen the best response.2then we can see which one of them will produce an int for the z coordinate too

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3There is a nice formula for number of representations of \(n\) as sum of two squares though

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[x^2+y^2 = 2016\] has no solutions over integers

dan815
 one year ago
Best ResponseYou've already chosen the best response.2this will give us a max bound on the number of integer solutions atleast

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3we can discard 2016 and work on smaller number 126 because both equations must have same number of solutions

dan815
 one year ago
Best ResponseYou've already chosen the best response.2how come we can do that?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(x^2+y^2+z^2=2016 = 16*126 \) 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
 one year ago
Best ResponseYou've already chosen the best response.2but we dont have to be able to factor 16 from all the squares

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3left hand side wont be divisible by \(16\) for all other cases, so we don't need to wry about them

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3for 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
 one year ago
Best ResponseYou've already chosen the best response.2what if we are dealing with cases where x y and z are all different mod and end up adding to 0 mod 16

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3thats 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
 one year ago
Best ResponseYou've already chosen the best response.2okk i see what u are saying

dan815
 one year ago
Best ResponseYou've already chosen the best response.2okay one more question, dont we have to consider the behavior in other mods too?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Nope, 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
 one year ago
Best ResponseYou've already chosen the best response.3this is same as considering even/odd cases right

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3If 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
 one year ago
Best ResponseYou've already chosen the best response.2can we continue the process then

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Any 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
 one year ago
Best ResponseYou've already chosen the best response.2and factor 3^2 and solve in mode 3?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3actually i was stuck with \[a^2+b^2+c^2 = 126\] mod 3 is not making it any simpler

dan815
 one year ago
Best ResponseYou've already chosen the best response.2the simplest one i can find

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Finding each solution gives us 3! = 6 solutions by permuting the ordered pairs

dan815
 one year ago
Best ResponseYou've already chosen the best response.2the choices are pretty low to pick from, maybe they expect brute force from here

dan815
 one year ago
Best ResponseYou've already chosen the best response.211 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
 one year ago
Best ResponseYou've already chosen the best response.2checking 9 81+45 45=6^2+3^2

dan815
 one year ago
Best ResponseYou've already chosen the best response.211,2,1 10,5,1 9,6,3 and so on

dan815
 one year ago
Best ResponseYou've already chosen the best response.2theres gotta be a better way than brute force

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3there 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
 one year ago
Best ResponseYou've already chosen the best response.2theres nothing with 8 and 7, so we are actually done here

dan815
 one year ago
Best ResponseYou've already chosen the best response.211,2,1 10,5,1 9,6,3 so 6*5*4 *3 right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.33*3! positive solitions 3*3!*2^3 integer solutions ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i 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
 one year ago
Best ResponseYou've already chosen the best response.2oh oops ya you are right, if i do this ill allow 11 and 11

dan815
 one year ago
Best ResponseYou've already chosen the best response.2(6*4*2) * 3 = 3*3! * 2^3 oh this works too

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yes 144 looks correct!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yups the answer is 144. You all are best mathematiciants :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you very much for your answer guys
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