Question

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!

Answer 1

just use gaussian elimination

Answer 2

oh.... I have bad eyes

Answer 3

I don't think so....

Answer 4

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

Answer 5

They call that a Diophantine equation, right?

Answer 6

You're probably off track!

Answer 7

lol

Answer 8

not very optimistic

Answer 9

I think the Z^4 is a hint

Answer 10

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Answer 11

pretty interesting question

Answer 12

wolfram's answer was not very comforting :/

Answer 13

lol yea turning test

Answer 14

let's write a Python script to solve this.

Answer 15

or maybe a calculator program?

Answer 16

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

Answer 17

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

Answer 18

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.

Answer 19

what exactly is the Diophantine equation?

Answer 20

but then we could write a dynamic programming solution!

Answer 21

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

Answer 22

http://en.wikipedia.org/wiki/Diophantine_equation

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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.

Answer 27

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

Answer 28

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

Answer 29

I like how the 4th dimension is time

Answer 30

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

Answer 31

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.

Answer 32

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

Answer 33

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 \((x,y,z,t)=(1,0,3,1)\).

Answer 34

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

Answer 35

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.

Answer 36

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

Answer 37

@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.

Answer 38

ok yea now i am getting it

Answer 39

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

Answer 40

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

Answer 41

Exactly!

Answer 42

Okay,
I'm getting the picture, thanks!

Answer 43

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

Answer 44

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).

Answer 45

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?

Answer 46

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

Answer 47

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

Answer 48

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?

Answer 49

I see it for t=z

Answer 50

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

Answer 51

x or z cannot be 0

Answer 52

Correct at Pax.

Answer 53

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.

Answer 54

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

Answer 55

the fact that the above has no integer solutions I mean

Answer 56

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

Answer 57

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

Answer 58

Likewise...

Answer 59

solutions*

Answer 60

Zarkon has it I think! :D

Answer 61

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

Answer 62

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

Answer 63

look above Pax, we have more than that

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

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 :)

Answer 68

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

Answer 69

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.

Answer 70

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 \[(x^2+2y^2)=11,\ (z^2+2t^2)=1\]

Answer 71

so just 4 solution sets...

Answer 72

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

Answer 73

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

Answer 74

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

Answer 75

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.