mathmate 3 years ago I am pretty tired, but would like to keep the party going, here's something to think about: Find all solutions in Z^4 for: xz-2yt = 3 xt + yz = 1 Good night!

1. agdgdgdgwngo

just use gaussian elimination

2. agdgdgdgwngo

oh.... I have bad eyes

3. TuringTest

I don't think so....

4. mathmate

There are four variables, only two equations, but the answer sets have to be integers.

5. TuringTest

They call that a Diophantine equation, right?

6. mathmate

You're probably off track!

7. agdgdgdgwngo

lol

8. TuringTest

not very optimistic

9. agdgdgdgwngo

I think the Z^4 is a hint

10. agdgdgdgwngo

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11. Akshay_Budhkar

pretty interesting question

12. TuringTest

wolfram's answer was not very comforting :/

13. Akshay_Budhkar

lol yea turning test

14. agdgdgdgwngo

let's write a Python script to solve this.

15. agdgdgdgwngo

or maybe a calculator program?

16. Mr.Math

If someone finds the answer using software, please don't post it yet!

17. agdgdgdgwngo

but if I could write such programs... then maybe I would be able to do this by hand :-P

18. TuringTest

I think these problems are harder to go about methodically. If it is a Diophantine equation, I seem to remember some theorem that says there is no way to know if they are generally solvable in a finite number of steps, which would make them hard to program I think.

19. Akshay_Budhkar

what exactly is the Diophantine equation?

20. agdgdgdgwngo

but then we could write a dynamic programming solution!

21. TuringTest

A Diophantine equation, if I remember right, is one that has integer solutions and more variables than equations. like all integer solutions to a=2b-3c

22. TuringTest
23. Akshay_Budhkar

um so how do we implement that here?? i am not getting any clue from that

24. TuringTest

beats me, I'm just thinking out loud...

25. Akshay_Budhkar

are u sure no complex number theroy can be helpful here ( just thinking out loud as well ) the Z^4 seems to be attracting complex theory somehow

26. TuringTest

Absolutely sounds appealing, but I know practically nothing of the subject. The Z^4 gets attention, but what's wrong with it exactly? There are four variables after all.

27. Mr.Math

He's looking for solutions over the integers, Z^4 stands for integers in order 4, for example (x,y,z,t).

28. Mr.Math

As we say R for real number x, and R^2 for the plane (x,y) and so on.

29. agdgdgdgwngo

I like how the 4th dimension is time

30. TuringTest

right, so it doesn't seem to have some trick answer based on the premise that the solutions are in z^4

31. Mr.Math

Actually it does. For example if you have can modify it under some conditions, say to $$xy+t=\frac{2}{3}$$, then you know there's no solution for this equation over the integers.

32. TuringTest

ahhh... thank you, that is insightful :)

33. Mr.Math

One solution that can be easily found is by taking $$y=0$$, (which is an integer). We will then have the system: $$xz=3$$ and $$xt=1$$ An obvious solution is $$x=t=1$$ and $$z=3$$. So we've got our first solution $$(x,y,z,t)=(1,0,3,1)$$.

34. Akshay_Budhkar

why did u take y=0? i didnt get that?

35. TuringTest

I'm thinking it's just to make it easier to "see" the solution, am I right? It presents fewer variables and reveals at least some of the solutions.

36. Mr.Math

For the case $$t=0$$, we will have $$y=z=1$$ and $$x=3$$. Thus the solution $$(3,1,10)$$.

37. Mr.Math

@Akshay: As turning said, we're dealing with four variables and with only two equations. So, it would make it easier to take the solutions case by case.

38. Akshay_Budhkar

ok yea now i am getting it

39. Mr.Math

A typo! The last solution is $$(3,1,1,0)$$.

40. TuringTest

I know, so taking x=0 gave yt=-3/2 so that means no integer solutions when x=0, right?

41. Mr.Math

Exactly!

42. TuringTest

Okay, I'm getting the picture, thanks!

43. TuringTest

so it looks like the only easy solutions are the ones you posted, but are there more?

44. Mr.Math

Similarly with z=0. Note here that xt=1, for example, has only the solution x=t=1 over Z. That means that the two solutions we found for y=0 and t=0, respectively, are the only solutions in these two cases (i.e No other integer solutions exist for y=0 or t=0).

45. TuringTest

Yes, I saw that and the fact that taking y=0 gives$x=\frac{1}{t}=\frac{3}{z}$and taking t=0 gives$z=\frac{3}{x}=\frac{1}{y}$means that the only possible integer solutions occurs when t=1 in the first set of equations and when y=1 in the second set of equations. So therefor there are no other solutions possible, yes?

46. Mr.Math

I can't say! We haven't showed that these are the only solutions, assuming that's true.

47. Mr.Math

I can say there are no solutions for $$x=y=z=t, x=y, \text{ or } t=z$$.

48. TuringTest

so we can't cover the cases for when z and/or x are not zero, but at least we have an element of the solution set down. ...what made you come to the above?

49. TuringTest

I see it for t=z

50. Mr.Math

Plug $$x=y=z=t$$, you get $$x^2-2x^2=3$$, which has no solution even over R.

51. PaxPolaris

x or z cannot be 0

52. Mr.Math

Correct at Pax.

53. Mr.Math

Also, no solutions for $$y=z$$, since it gives the system: $$xy-2yt=3$$ and $$xt+y^2=1$$, and this system has no integer solutions.

54. TuringTest

I can't see that by looking at it, maybe I'm missing it.

55. TuringTest

the fact that the above has no integer solutions I mean

56. Mr.Math

Actually it does, but with the condition that t=0, and this would give us a solution that we have already found.

57. Mr.Math

For some reason, I feel those two solutions are the only solution. But I can't prove it!

58. TuringTest

Likewise...

59. Mr.Math

solutions*

60. Mr.Math

Zarkon has it I think! :D

61. TuringTest

We've at least proved that those are the only solution in the case that either y or t are zero.

62. PaxPolaris

all I got so far is the x & z have to be odd (y,t!=0):$yt={xz-3 \over 2}$

63. TuringTest

look above Pax, we have more than that

64. Mr.Math

We've also proved, for several cases, that no integer solutions exist. Yet, that is not enough.

65. Zarkon

xz-2yt = 3 xt + yz = 1 suppose y>1 and z>0 then xt=1-yz<0 so xt<0 so (x<0 and t>0) or (x>0 and t<0) if x<0 and t>0 then from xz-2yt = 3 =>xz=3+2yt ty>0 so 3+yt>0 and thus xz>0 thus x>0 a contradiction if x>0 and t<0 then 2yt<-3 and thus 3+2yt<0 so xz<0 but x,z>0 a contradiction thus no solution for y>1 and z>0

66. Zarkon

I believe I have a similar argument for y<0 and z>0

67. TuringTest

I'm trying to see just how large the ramifications of that are, but I'm too tired, so I'm going to bed. *bookmarked Good night :)

68. Zarkon

note in the above argument that y>1 is the same as $$y\ge 2$$ and z>0 is the same as $$z\ge 1$$

69. Zarkon

ok... xz-2yt = 3 xt + yz = 1 $11=3^2+2\cdot 1^2=(xz - 2yt)^2 + 2(xt + yz)^2 = (x^2 + 2z^2)(y^2 + 2t^2)$ so $(x^2 + 2z^2)=11,(y^2 + 2t^2)=1$ or $(x^2 + 2z^2)=1,(y^2 + 2t^2)=11$ now solve ... very easy from here.

70. PaxPolaris

Cool, but you got a slight typo: $(xz−2yt)^2+2(xt+yz)^2=x^2z^2+4y^2t^2+2x^2t^2+2y^2z^2$$=(x^2+2y^2)(z^2+2t^2)=11$ $(x^2+2y^2)=1,\ (z^2+2t^2)=11$ or $(x^2+2y^2)=11,\ (z^2+2t^2)=1$

71. PaxPolaris

so just 4 solution sets...

72. mathmate

Wow, you guys are amaaaaazing!!!! I'm totally impressed.

73. Mr.Math

Not that amazing! I was stupid enough to not notice that x=z=-1 is an integer solution for xz=1. ;)

74. Mr.Math

It was a nice problem. Of course, only Zarkon proved that only four solutions are there! :D

75. mathmate

To sum up, the solution set for (x^2+2y^2)=1 will have integer solutions if y=0, and x=+/-1. (z^2+2t^2)=11 will have integer solutions if z=+/-3 and t=+/-1. However, to satisfy the original equations, the only four solution sets are, in (x,y,z,t) space: (1,0,3,1), (-1,0,-3,-1), (3,1,1,0) and (-3,-1,-1,0) Thank you everyone who participated, and a special appreciation for Zarkon.