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mathmate
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!
just use gaussian elimination
oh.... I have bad eyes
I don't think so....
There are four variables, only two equations, but the answer sets have to be integers.
They call that a Diophantine equation, right?
You're probably off track!
I think the Z^4 is a hint
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pretty interesting question
wolfram's answer was not very comforting :/
lol yea turning test
let's write a Python script to solve this.
or maybe a calculator program?
If someone finds the answer using software, please don't post it yet!
but if I could write such programs... then maybe I would be able to do this by hand :-P
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.
what exactly is the Diophantine equation?
but then we could write a dynamic programming solution!
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
um so how do we implement that here?? i am not getting any clue from that
beats me, I'm just thinking out loud...
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
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.
He's looking for solutions over the integers, Z^4 stands for integers in order 4, for example (x,y,z,t).
As we say R for real number x, and R^2 for the plane (x,y) and so on.
I like how the 4th dimension is time
right, so it doesn't seem to have some trick answer based on the premise that the solutions are in z^4
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.
ahhh... thank you, that is insightful :)
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 \((x,y,z,t)=(1,0,3,1)\).
why did u take y=0? i didnt get that?
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.
For the case \(t=0\), we will have \(y=z=1\) and \(x=3\). Thus the solution \((3,1,10)\).
@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.
ok yea now i am getting it
A typo! The last solution is \((3,1,1,0)\).
I know, so taking x=0 gave yt=-3/2 so that means no integer solutions when x=0, right?
Okay, I'm getting the picture, thanks!
so it looks like the only easy solutions are the ones you posted, but are there more?
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).
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?
I can't say! We haven't showed that these are the only solutions, assuming that's true.
I can say there are no solutions for \(x=y=z=t, x=y, \text{ or } t=z\).
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?
Plug \(x=y=z=t\), you get \(x^2-2x^2=3\), which has no solution even over R.
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.
I can't see that by looking at it, maybe I'm missing it.
the fact that the above has no integer solutions I mean
Actually it does, but with the condition that t=0, and this would give us a solution that we have already found.
For some reason, I feel those two solutions are the only solution. But I can't prove it!
Zarkon has it I think! :D
We've at least proved that those are the only solution in the case that either y or t are zero.
all I got so far is the x & z have to be odd (y,t!=0):\[yt={xz-3 \over 2}\]
look above Pax, we have more than that
We've also proved, for several cases, that no integer solutions exist. Yet, that is not enough.
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
I believe I have a similar argument for y<0 and z>0
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 :)
note in the above argument that y>1 is the same as \(y\ge 2\) and z>0 is the same as \(z\ge 1\)
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.
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 \[(x^2+2y^2)=11,\ (z^2+2t^2)=1\]
so just 4 solution sets...
Wow, you guys are amaaaaazing!!!! I'm totally impressed.
Not that amazing! I was stupid enough to not notice that x=z=-1 is an integer solution for xz=1. ;)
It was a nice problem. Of course, only Zarkon proved that only four solutions are there! :D
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.