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

1. ganeshie8

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

2. 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 $$x = 4a$$, $$y=4b$$, $$z=4c$$

3. anonymous

apparently gauss did it over 200 years ago but hmm

4. 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..

5. anonymous

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

6. ganeshie8

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

7. dan815

|dw:1436766690043:dw|

8. dan815

|dw:1436766915671:dw|

9. dan815

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10. dan815

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

11. dan815

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

12. ganeshie8

There is a nice formula for number of representations of $$n$$ as sum of two squares though

13. ganeshie8

$x^2+y^2 = 2016$ has no solutions over integers

14. dan815

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15. dan815

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

16. ganeshie8

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

17. dan815

how come we can do that?

18. 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 $$x^2+y^2+z^2 = 16(a^2+b^2+c^2) = 2016 \implies a^2+b^2+c^2 = 126$$

19. dan815

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

20. ganeshie8

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

21. 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

22. 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

23. 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 $$\equiv 0\pmod{4}$$

24. dan815

okk i see what u are saying

25. dan815

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

26. 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

27. ganeshie8

this is same as considering even/odd cases right

28. 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

29. dan815

ok isee

30. dan815

can we continue the process then

31. 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 $$X\equiv Y\equiv Z\equiv 0$$

32. dan815

and factor 3^2 and solve in mode 3?

33. ganeshie8

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

34. dan815

11,2,1

35. dan815

the simplest one i can find

36. ganeshie8

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

37. dan815

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

38. ganeshie8

yeah

39. 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

40. dan815

oh wait 25+1

41. dan815

10 is good too

42. dan815

nothing else

43. dan815

11,2,1 10,5,1

44. dan815

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

45. dan815

9,6,3

46. ganeshie8

thats all

47. dan815

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

48. dan815

theres gotta be a better way than brute force

49. 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

50. dan815

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

51. dan815

since it will repeat

52. dan815

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

53. ganeshie8

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

54. 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

55. dan815

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

56. dan815

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

57. dan815

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58. ganeshie8

yes 144 looks correct!

59. anonymous

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

60. anonymous

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