## mathmate Group Title 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! 2 years ago 2 years ago

1. agdgdgdgwngo Group Title

just use gaussian elimination

2. agdgdgdgwngo Group Title

oh.... I have bad eyes

3. TuringTest Group Title

I don't think so....

4. mathmate Group Title

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

5. TuringTest Group Title

They call that a Diophantine equation, right?

6. mathmate Group Title

You're probably off track!

7. agdgdgdgwngo Group Title

lol

8. TuringTest Group Title

not very optimistic

9. agdgdgdgwngo Group Title

I think the Z^4 is a hint

10. agdgdgdgwngo Group Title

\begin{array}l\color{#FF0000}{\text{|}}\color{#FF7F00}{\text{|}}\color{#FFE600}{\text{|}}\color{#00FF00}{\text{|}}\color{#0000FF}{\text{|}}\color{#6600FF}{\text{|}}\color{#8B00FF}{\text{|}}\color{#FF0000}{\text{|}}\color{#FF7F00}{\text{|}}\color{#FFE600}{\text{|}}\color{#00FF00}{\text{|}}\color{#0000FF}{\text{|}}\color{#6600FF}{\text{|}}\color{#8B00FF}{\text{|}}\color{#FF0000}{\text{|}}\color{#FF7F00}{\text{|}}\color{#FFE600}{\text{|}}\color{#00FF00}{\text{|}}\color{#0000FF}{\text{|}}\color{#6600FF}{\text{|}}\color{#8B00FF}{\text{|}}\color{#FF0000}{\text{|}}\color{#FF7F00}{\text{|}}\color{#FFE600}{\text{|}}\color{#00FF00}{\text{|}}\color{#0000FF}{\text{|}}\color{#6600FF}{\text{|}}\color{#8B00FF}{\text{|}}\color{#FF0000}{\text{|}}\color{#FF7F00}{\text{|}}\color{#FFE600}{\text{|}}\color{#00FF00}{\text{|}}\color{#0000FF}{\text{|}}\color{#6600FF}{\text{|}}\color{#8B00FF}{\text{|}}\color{#FF0000}{\text{|}}\color{#FF7F00}{\text{|}}\color{#FFE600}{\text{|}}\color{#00FF00}{\text{|}}\color{#0000FF}{\text{|}}\end{array}

11. Akshay_Budhkar Group Title

pretty interesting question

12. TuringTest Group Title

wolfram's answer was not very comforting :/

13. Akshay_Budhkar Group Title

lol yea turning test

14. agdgdgdgwngo Group Title

let's write a Python script to solve this.

15. agdgdgdgwngo Group Title

or maybe a calculator program?

16. Mr.Math Group Title

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

17. agdgdgdgwngo Group Title

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

18. TuringTest Group Title

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 Group Title

what exactly is the Diophantine equation?

20. agdgdgdgwngo Group Title

but then we could write a dynamic programming solution!

21. TuringTest Group Title

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 Group Title
23. Akshay_Budhkar Group Title

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

24. TuringTest Group Title

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

25. Akshay_Budhkar Group Title

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 Group Title

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 Group Title

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 Group Title

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

29. agdgdgdgwngo Group Title

I like how the 4th dimension is time

30. TuringTest Group Title

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 Group Title

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 Group Title

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

33. Mr.Math Group Title

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 Group Title

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

35. TuringTest Group Title

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 Group Title

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

37. Mr.Math Group Title

@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 Group Title

ok yea now i am getting it

39. Mr.Math Group Title

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

40. TuringTest Group Title

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

41. Mr.Math Group Title

Exactly!

42. TuringTest Group Title

Okay, I'm getting the picture, thanks!

43. TuringTest Group Title

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

44. Mr.Math Group Title

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 Group Title

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 Group Title

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

47. Mr.Math Group Title

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

48. TuringTest Group Title

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 Group Title

I see it for t=z

50. Mr.Math Group Title

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

51. PaxPolaris Group Title

x or z cannot be 0

52. Mr.Math Group Title

Correct at Pax.

53. Mr.Math Group Title

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 Group Title

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

55. TuringTest Group Title

the fact that the above has no integer solutions I mean

56. Mr.Math Group Title

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 Group Title

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

58. TuringTest Group Title

Likewise...

59. Mr.Math Group Title

solutions*

60. Mr.Math Group Title

Zarkon has it I think! :D

61. TuringTest Group Title

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

62. PaxPolaris Group Title

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

63. TuringTest Group Title

look above Pax, we have more than that

64. Mr.Math Group Title

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

65. Zarkon Group Title

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 Group Title

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

67. TuringTest Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

so just 4 solution sets...

72. mathmate Group Title

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

73. Mr.Math Group Title

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

74. Mr.Math Group Title

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

75. mathmate Group Title

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.