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!

- mathmate

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- anonymous

just use gaussian elimination

- anonymous

oh.... I have bad eyes

- TuringTest

I don't think so....

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## More answers

- mathmate

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

- TuringTest

They call that a Diophantine equation, right?

- mathmate

You're probably off track!

- anonymous

lol

- TuringTest

not very optimistic

- anonymous

I think the Z^4 is a hint

- anonymous

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- Akshay_Budhkar

pretty interesting question

- TuringTest

wolfram's answer was not very comforting :/

- Akshay_Budhkar

lol yea turning test

- anonymous

let's write a Python script to solve this.

- anonymous

or maybe a calculator program?

- Mr.Math

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

- anonymous

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

- 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.

- Akshay_Budhkar

what exactly is the Diophantine equation?

- anonymous

but then we could write a dynamic programming solution!

- 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

- TuringTest

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

- Akshay_Budhkar

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

- TuringTest

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

- 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

- 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.

- Mr.Math

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

- Mr.Math

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

- anonymous

I like how the 4th dimension is time

- TuringTest

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

- 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.

- TuringTest

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

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

- Akshay_Budhkar

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

- 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.

- Mr.Math

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

- 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.

- Akshay_Budhkar

ok yea now i am getting it

- Mr.Math

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

- TuringTest

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

- Mr.Math

Exactly!

- TuringTest

Okay,
I'm getting the picture, thanks!

- TuringTest

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

- 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).

- 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?

- Mr.Math

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

- Mr.Math

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

- 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?

- TuringTest

I see it for t=z

- Mr.Math

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

- PaxPolaris

x or z cannot be 0

- Mr.Math

Correct at Pax.

- 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.

- TuringTest

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

- TuringTest

the fact that the above has no integer solutions I mean

- Mr.Math

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

- Mr.Math

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

- TuringTest

Likewise...

- Mr.Math

solutions*

- Mr.Math

Zarkon has it I think! :D

- TuringTest

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

- PaxPolaris

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

- TuringTest

look above Pax, we have more than that

- Mr.Math

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

- 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

- Zarkon

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

- 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 :)

- Zarkon

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

- 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.

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

- PaxPolaris

so just 4 solution sets...

- mathmate

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

- Mr.Math

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

- Mr.Math

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

- 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.

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