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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:
xz2yt = 3
xt + yz = 1
Good night!
 2 years ago
 2 years ago
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: xz2yt = 3 xt + yz = 1 Good night!
 2 years ago
 2 years ago

This Question is Closed

agdgdgdgwngo Group TitleBest ResponseYou've already chosen the best response.0
just use gaussian elimination
 2 years ago

agdgdgdgwngo Group TitleBest ResponseYou've already chosen the best response.0
oh.... I have bad eyes
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I don't think so....
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
There are four variables, only two equations, but the answer sets have to be integers.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
They call that a Diophantine equation, right?
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
You're probably off track!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
not very optimistic
 2 years ago

agdgdgdgwngo Group TitleBest ResponseYou've already chosen the best response.0
I think the Z^4 is a hint
 2 years ago

agdgdgdgwngo Group TitleBest ResponseYou've already chosen the best response.0
\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}
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
pretty interesting question
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
wolfram's answer was not very comforting :/
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
lol yea turning test
 2 years ago

agdgdgdgwngo Group TitleBest ResponseYou've already chosen the best response.0
let's write a Python script to solve this.
 2 years ago

agdgdgdgwngo Group TitleBest ResponseYou've already chosen the best response.0
or maybe a calculator program?
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
If someone finds the answer using software, please don't post it yet!
 2 years ago

agdgdgdgwngo Group TitleBest ResponseYou've already chosen the best response.0
but if I could write such programs... then maybe I would be able to do this by hand :P
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
what exactly is the Diophantine equation?
 2 years ago

agdgdgdgwngo Group TitleBest ResponseYou've already chosen the best response.0
but then we could write a dynamic programming solution!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
A Diophantine equation, if I remember right, is one that has integer solutions and more variables than equations. like all integer solutions to a=2b3c
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
http://en.wikipedia.org/wiki/Diophantine_equation
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
um so how do we implement that here?? i am not getting any clue from that
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
beats me, I'm just thinking out loud...
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
He's looking for solutions over the integers, Z^4 stands for integers in order 4, for example (x,y,z,t).
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
As we say R for real number x, and R^2 for the plane (x,y) and so on.
 2 years ago

agdgdgdgwngo Group TitleBest ResponseYou've already chosen the best response.0
I like how the 4th dimension is time
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
right, so it doesn't seem to have some trick answer based on the premise that the solutions are in z^4
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
ahhh... thank you, that is insightful :)
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
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 \((x,y,z,t)=(1,0,3,1)\).
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
why did u take y=0? i didnt get that?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
For the case \(t=0\), we will have \(y=z=1\) and \(x=3\). Thus the solution \((3,1,10)\).
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
@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.
 2 years ago

Akshay_Budhkar Group TitleBest ResponseYou've already chosen the best response.0
ok yea now i am getting it
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
A typo! The last solution is \((3,1,1,0)\).
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I know, so taking x=0 gave yt=3/2 so that means no integer solutions when x=0, right?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
Okay, I'm getting the picture, thanks!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
so it looks like the only easy solutions are the ones you posted, but are there more?
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
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).
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.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?
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
I can't say! We haven't showed that these are the only solutions, assuming that's true.
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
I can say there are no solutions for \(x=y=z=t, x=y, \text{ or } t=z\).
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I see it for t=z
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
Plug \(x=y=z=t\), you get \(x^22x^2=3\), which has no solution even over R.
 2 years ago

PaxPolaris Group TitleBest ResponseYou've already chosen the best response.0
x or z cannot be 0
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
Correct at Pax.
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
Also, no solutions for \(y=z\), since it gives the system: \(xy2yt=3\) and \(xt+y^2=1\), and this system has no integer solutions.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
I can't see that by looking at it, maybe I'm missing it.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
the fact that the above has no integer solutions I mean
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
Actually it does, but with the condition that t=0, and this would give us a solution that we have already found.
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
For some reason, I feel those two solutions are the only solution. But I can't prove it!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
Likewise...
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
Zarkon has it I think! :D
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
We've at least proved that those are the only solution in the case that either y or t are zero.
 2 years ago

PaxPolaris Group TitleBest ResponseYou've already chosen the best response.0
all I got so far is the x & z have to be odd (y,t!=0):\[yt={xz3 \over 2}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
look above Pax, we have more than that
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
We've also proved, for several cases, that no integer solutions exist. Yet, that is not enough.
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.6
xz2yt = 3 xt + yz = 1 suppose y>1 and z>0 then xt=1yz<0 so xt<0 so (x<0 and t>0) or (x>0 and t<0) if x<0 and t>0 then from xz2yt = 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
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.6
I believe I have a similar argument for y<0 and z>0
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.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 :)
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.6
note in the above argument that y>1 is the same as \(y\ge 2\) and z>0 is the same as \(z\ge 1\)
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.6
ok... xz2yt = 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.
 2 years ago

PaxPolaris Group TitleBest ResponseYou've already chosen the best response.0
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 \[(x^2+2y^2)=11,\ (z^2+2t^2)=1\]
 2 years ago

PaxPolaris Group TitleBest ResponseYou've already chosen the best response.0
so just 4 solution sets...
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
Wow, you guys are amaaaaazing!!!! I'm totally impressed.
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
Not that amazing! I was stupid enough to not notice that x=z=1 is an integer solution for xz=1. ;)
 2 years ago

Mr.Math Group TitleBest ResponseYou've already chosen the best response.0
It was a nice problem. Of course, only Zarkon proved that only four solutions are there! :D
 2 years ago

mathmate Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago
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