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

- jamiebookeater

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.