How many solutions which integers such that x^2 + y^2 + z^2 = 2016

- anonymous

How many solutions which integers such that x^2 + y^2 + z^2 = 2016

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- ganeshie8

I think it is sufficient to consider solutions in positive integers first, then we can build the complete solutions by permutations.

- 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

apparently gauss did it over 200 years ago but hmm

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## More answers

- 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

actually, he did it for square free numbers, so I guess we're screwed

- ganeshie8

looks interesting
http://gyazo.com/301173b00539b4577e227beafd0b4281

- dan815

|dw:1436766690043:dw|

- dan815

|dw:1436766915671:dw|

- dan815

|dw:1436767224913:dw|

- dan815

lets start by considering the total number of ordered pairs of inteegers in this arc

- dan815

then we can see which one of them will produce an int for the z coordinate too

- ganeshie8

There is a nice formula for number of representations of \(n\) as sum of two squares though

- ganeshie8

\[x^2+y^2 = 2016\]
has no solutions over integers

- dan815

|dw:1436767447283:dw|

- dan815

this will give us a max bound on the number of integer solutions atleast

- ganeshie8

we can discard 2016 and work on smaller number 126 because both equations must have same number of solutions

- dan815

how come we can do that?

- 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

but we dont have to be able to factor 16 from all the squares

- ganeshie8

left hand side wont be divisible by \(16\) for all other cases, so we don't need to wry about them

- 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

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

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

okk i see what u are saying

- dan815

okay one more question, dont we have to consider the behavior in other mods too?

- 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

this is same as considering even/odd cases right

- 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

ok isee

- dan815

can we continue the process then

- 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

and factor 3^2 and solve in mode 3?

- ganeshie8

actually i was stuck with
\[a^2+b^2+c^2 = 126\]
mod 3 is not making it any simpler

- dan815

11,2,1

- dan815

the simplest one i can find

- ganeshie8

Finding each solution gives us 3! = 6 solutions by permuting the ordered pairs

- dan815

the choices are pretty low to pick from, maybe they expect brute force from here

- ganeshie8

yeah

- 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

oh wait 25+1

- dan815

10 is good too

- dan815

nothing else

- dan815

11,2,1
10,5,1

- dan815

checking 9
81+45
45=6^2+3^2

- dan815

9,6,3

- ganeshie8

thats all

- dan815

11,2,1
10,5,1
9,6,3
and so on

- dan815

theres gotta be a better way than brute force

- 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

theres nothing with 8 and 7, so we are actually done here

- dan815

since it will repeat

- dan815

11,2,1
10,5,1
9,6,3
so 6*5*4 *3 right?

- ganeshie8

3*3! positive solitions
3*3!*2^3 integer solutions ?

- 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

oh oops ya you are right, if i do this ill allow -11 and 11

- dan815

(6*4*2) * 3 = 3*3! * 2^3 oh this works too

- dan815

|dw:1436769950610:dw|

- ganeshie8

yes 144 looks correct!

- anonymous

Yups the answer is 144. You all are best mathematiciants :)

- anonymous

Thank you very much for your answer guys

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