Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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!

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

just use gaussian elimination
oh.... I have bad eyes
I don't think so....

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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!
lol
not very optimistic
I think the Z^4 is a hint
\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}
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
http://en.wikipedia.org/wiki/Diophantine_equation
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?
Exactly!
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?
I see it for t=z
Plug \(x=y=z=t\), you get \(x^2-2x^2=3\), which has no solution even over R.
x or z cannot be 0
Correct at Pax.
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!
Likewise...
solutions*
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question